A small block of mass m = 200 g is released from rest at point Ⓐ along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P7.45). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point Ⓐ relative to point Ⓑ. (b) the kinetic energy of the block at point Ⓑ, (c) its speed at point Ⓑ, and (d) its kinetic energy and the potential energy when the block is at point Ⓒ. Figure P7.45 Problems 45 and 46.

Question

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Answer 1

Textbook Question

Chapter 7, Problem 45P

A small block of mass m = 200 g is released from rest at point Ⓐ along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P7.45). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point Ⓐ relative to point Ⓑ. (b) the kinetic energy of the block at point Ⓑ, (c) its speed at point Ⓑ, and (d) its kinetic energy and the potential energy when the block is at point Ⓒ.

Chapter 7, Problem 45P, A small block of mass m = 200 g is released from rest at point along the horizontal diameter on the

Figure P7.45 Problems 45 and 46.

(a)

Expert Solution

To determine

Gravitational potential energy.

Answer to Problem 45P

The gravitational potential energy is 0.588 J_.

Explanation of Solution

Write down the equation for gravitational potential energy

UB=mgR (I)

Here UB is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and R is the distance.

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for R in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(b)

Expert Solution

To determine

Kinetic energy of the block.

Answer to Problem 45P

The kinetic energy is 0.588 J_.

Explanation of Solution

Write the law of conservation of energy.

KA+UA=KB+UB (II)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KB is the kinetic energy at point B and UB is the potential energy at point B.

At point A, kinetic energy is zero.

At point B, potential energy is zero.

Therefore,

KB=UA (III)

Substitute (I) in (III)

KB=mgR (IV)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for h in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(c)

Expert Solution

To determine

Velocity at point B.

Answer to Problem 45P

Velocity is 2.42 m/s_

Explanation of Solution

Write down the equation for kinetic energy.

KB=12mvB2 (V)

Here KB is the kinetic energy at point B, m is the mass of the block, vB is the velocity

Rewrite (V) in terms of vB.

vB=2KBm (VI)

Conclusion:

Substitute 0.588 J for KB and 200 g for m in (VI).

vB=2(0.588 J)(200 g(10−3 kg1 g))=2(0.588 J)(200×10−3 kg)=2.42 m/s

Velocity is 2.42 m/s_

(d)

Expert Solution

To determine

Kinetic energy and potential energy at point C.

Answer to Problem 45P

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

Explanation of Solution

Write down the equation for gravitational potential energy

UC=mghC (VII)

Here UC is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and hC is the height at point C.

Write the law of conservation of energy.

KA+UA=KC+UC (VIII)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KC is the kinetic energy at point C and UC is the potential energy at point C.

Rewrite (VIII)

KC=KA+UA−UC

At point A, kinetic energy is zero.

Therefore,

KC=UA−UC (IX)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 20 cm for hc in (VII).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(20 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(20×10−2 m)=0.392 J

Substitute 0.588 J for UA and 0.392 J for UC in (IX)

KC=0.588 J−0.392 J=0.196 J

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

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Answer 2

Textbook Question

Chapter 7, Problem 45P

A small block of mass m = 200 g is released from rest at point Ⓐ along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P7.45). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point Ⓐ relative to point Ⓑ. (b) the kinetic energy of the block at point Ⓑ, (c) its speed at point Ⓑ, and (d) its kinetic energy and the potential energy when the block is at point Ⓒ.

Chapter 7, Problem 45P, A small block of mass m = 200 g is released from rest at point along the horizontal diameter on the

Figure P7.45 Problems 45 and 46.

(a)

Expert Solution

To determine

Gravitational potential energy.

Answer to Problem 45P

The gravitational potential energy is 0.588 J_.

Explanation of Solution

Write down the equation for gravitational potential energy

UB=mgR (I)

Here UB is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and R is the distance.

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for R in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(b)

Expert Solution

To determine

Kinetic energy of the block.

Answer to Problem 45P

The kinetic energy is 0.588 J_.

Explanation of Solution

Write the law of conservation of energy.

KA+UA=KB+UB (II)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KB is the kinetic energy at point B and UB is the potential energy at point B.

At point A, kinetic energy is zero.

At point B, potential energy is zero.

Therefore,

KB=UA (III)

Substitute (I) in (III)

KB=mgR (IV)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for h in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(c)

Expert Solution

To determine

Velocity at point B.

Answer to Problem 45P

Velocity is 2.42 m/s_

Explanation of Solution

Write down the equation for kinetic energy.

KB=12mvB2 (V)

Here KB is the kinetic energy at point B, m is the mass of the block, vB is the velocity

Rewrite (V) in terms of vB.

vB=2KBm (VI)

Conclusion:

Substitute 0.588 J for KB and 200 g for m in (VI).

vB=2(0.588 J)(200 g(10−3 kg1 g))=2(0.588 J)(200×10−3 kg)=2.42 m/s

Velocity is 2.42 m/s_

(d)

Expert Solution

To determine

Kinetic energy and potential energy at point C.

Answer to Problem 45P

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

Explanation of Solution

Write down the equation for gravitational potential energy

UC=mghC (VII)

Here UC is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and hC is the height at point C.

Write the law of conservation of energy.

KA+UA=KC+UC (VIII)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KC is the kinetic energy at point C and UC is the potential energy at point C.

Rewrite (VIII)

KC=KA+UA−UC

At point A, kinetic energy is zero.

Therefore,

KC=UA−UC (IX)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 20 cm for hc in (VII).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(20 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(20×10−2 m)=0.392 J

Substitute 0.588 J for UA and 0.392 J for UC in (IX)

KC=0.588 J−0.392 J=0.196 J

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

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Answer 3

Textbook Question

Chapter 7, Problem 45P

A small block of mass m = 200 g is released from rest at point Ⓐ along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P7.45). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point Ⓐ relative to point Ⓑ. (b) the kinetic energy of the block at point Ⓑ, (c) its speed at point Ⓑ, and (d) its kinetic energy and the potential energy when the block is at point Ⓒ.

Chapter 7, Problem 45P, A small block of mass m = 200 g is released from rest at point along the horizontal diameter on the

Figure P7.45 Problems 45 and 46.

(a)

Expert Solution

To determine

Gravitational potential energy.

Answer to Problem 45P

The gravitational potential energy is 0.588 J_.

Explanation of Solution

Write down the equation for gravitational potential energy

UB=mgR (I)

Here UB is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and R is the distance.

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for R in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(b)

Expert Solution

To determine

Kinetic energy of the block.

Answer to Problem 45P

The kinetic energy is 0.588 J_.

Explanation of Solution

Write the law of conservation of energy.

KA+UA=KB+UB (II)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KB is the kinetic energy at point B and UB is the potential energy at point B.

At point A, kinetic energy is zero.

At point B, potential energy is zero.

Therefore,

KB=UA (III)

Substitute (I) in (III)

KB=mgR (IV)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for h in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(c)

Expert Solution

To determine

Velocity at point B.

Answer to Problem 45P

Velocity is 2.42 m/s_

Explanation of Solution

Write down the equation for kinetic energy.

KB=12mvB2 (V)

Here KB is the kinetic energy at point B, m is the mass of the block, vB is the velocity

Rewrite (V) in terms of vB.

vB=2KBm (VI)

Conclusion:

Substitute 0.588 J for KB and 200 g for m in (VI).

vB=2(0.588 J)(200 g(10−3 kg1 g))=2(0.588 J)(200×10−3 kg)=2.42 m/s

Velocity is 2.42 m/s_

(d)

Expert Solution

To determine

Kinetic energy and potential energy at point C.

Answer to Problem 45P

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

Explanation of Solution

Write down the equation for gravitational potential energy

UC=mghC (VII)

Here UC is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and hC is the height at point C.

Write the law of conservation of energy.

KA+UA=KC+UC (VIII)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KC is the kinetic energy at point C and UC is the potential energy at point C.

Rewrite (VIII)

KC=KA+UA−UC

At point A, kinetic energy is zero.

Therefore,

KC=UA−UC (IX)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 20 cm for hc in (VII).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(20 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(20×10−2 m)=0.392 J

Substitute 0.588 J for UA and 0.392 J for UC in (IX)

KC=0.588 J−0.392 J=0.196 J

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

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Answer 4

Textbook Question

Chapter 7, Problem 45P

A small block of mass m = 200 g is released from rest at point Ⓐ along the horizontal diameter on the inside of a frictionless, hemispherical bowl of radius R = 30.0 cm (Fig. P7.45). Calculate (a) the gravitational potential energy of the block-Earth system when the block is at point Ⓐ relative to point Ⓑ. (b) the kinetic energy of the block at point Ⓑ, (c) its speed at point Ⓑ, and (d) its kinetic energy and the potential energy when the block is at point Ⓒ.

Chapter 7, Problem 45P, A small block of mass m = 200 g is released from rest at point along the horizontal diameter on the

Figure P7.45 Problems 45 and 46.

(a)

Expert Solution

To determine

Gravitational potential energy.

Answer to Problem 45P

The gravitational potential energy is 0.588 J_.

Explanation of Solution

Write down the equation for gravitational potential energy

UB=mgR (I)

Here UB is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and R is the distance.

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for R in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(b)

Expert Solution

To determine

Kinetic energy of the block.

Answer to Problem 45P

The kinetic energy is 0.588 J_.

Explanation of Solution

Write the law of conservation of energy.

KA+UA=KB+UB (II)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KB is the kinetic energy at point B and UB is the potential energy at point B.

At point A, kinetic energy is zero.

At point B, potential energy is zero.

Therefore,

KB=UA (III)

Substitute (I) in (III)

KB=mgR (IV)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for h in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(c)

Expert Solution

To determine

Velocity at point B.

Answer to Problem 45P

Velocity is 2.42 m/s_

Explanation of Solution

Write down the equation for kinetic energy.

KB=12mvB2 (V)

Here KB is the kinetic energy at point B, m is the mass of the block, vB is the velocity

Rewrite (V) in terms of vB.

vB=2KBm (VI)

Conclusion:

Substitute 0.588 J for KB and 200 g for m in (VI).

vB=2(0.588 J)(200 g(10−3 kg1 g))=2(0.588 J)(200×10−3 kg)=2.42 m/s

Velocity is 2.42 m/s_

(d)

Expert Solution

To determine

Kinetic energy and potential energy at point C.

Answer to Problem 45P

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

Explanation of Solution

Write down the equation for gravitational potential energy

UC=mghC (VII)

Here UC is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and hC is the height at point C.

Write the law of conservation of energy.

KA+UA=KC+UC (VIII)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KC is the kinetic energy at point C and UC is the potential energy at point C.

Rewrite (VIII)

KC=KA+UA−UC

At point A, kinetic energy is zero.

Therefore,

KC=UA−UC (IX)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 20 cm for hc in (VII).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(20 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(20×10−2 m)=0.392 J

Substitute 0.588 J for UA and 0.392 J for UC in (IX)

KC=0.588 J−0.392 J=0.196 J

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

Gravitational potential energy.

Answer to Problem 45P

The gravitational potential energy is 0.588 J_.

Explanation of Solution

Write down the equation for gravitational potential energy

UB=mgR (I)

Here UB is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and R is the distance.

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for R in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(b)

Expert Solution

To determine

Kinetic energy of the block.

Answer to Problem 45P

The kinetic energy is 0.588 J_.

Explanation of Solution

Write the law of conservation of energy.

KA+UA=KB+UB (II)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KB is the kinetic energy at point B and UB is the potential energy at point B.

At point A, kinetic energy is zero.

At point B, potential energy is zero.

Therefore,

KB=UA (III)

Substitute (I) in (III)

KB=mgR (IV)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for h in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

(c)

Expert Solution

To determine

Velocity at point B.

Answer to Problem 45P

Velocity is 2.42 m/s_

Explanation of Solution

Write down the equation for kinetic energy.

KB=12mvB2 (V)

Here KB is the kinetic energy at point B, m is the mass of the block, vB is the velocity

Rewrite (V) in terms of vB.

vB=2KBm (VI)

Conclusion:

Substitute 0.588 J for KB and 200 g for m in (VI).

vB=2(0.588 J)(200 g(10−3 kg1 g))=2(0.588 J)(200×10−3 kg)=2.42 m/s

Velocity is 2.42 m/s_

(d)

Expert Solution

To determine

Kinetic energy and potential energy at point C.

Answer to Problem 45P

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

Explanation of Solution

Write down the equation for gravitational potential energy

UC=mghC (VII)

Here UC is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and hC is the height at point C.

Write the law of conservation of energy.

KA+UA=KC+UC (VIII)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KC is the kinetic energy at point C and UC is the potential energy at point C.

Rewrite (VIII)

KC=KA+UA−UC

At point A, kinetic energy is zero.

Therefore,

KC=UA−UC (IX)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 20 cm for hc in (VII).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(20 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(20×10−2 m)=0.392 J

Substitute 0.588 J for UA and 0.392 J for UC in (IX)

KC=0.588 J−0.392 J=0.196 J

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

Answer 8

(a)

Expert Solution

To determine

Gravitational potential energy.

Answer to Problem 45P

The gravitational potential energy is 0.588 J_.

Explanation of Solution

Write down the equation for gravitational potential energy

UB=mgR (I)

Here UB is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and R is the distance.

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for R in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Gravitational potential energy.

Answer 13

Answer to Problem 45P

The gravitational potential energy is 0.588 J_.

Answer 14

Explanation of Solution

Write down the equation for gravitational potential energy

UB=mgR (I)

Here UB is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and R is the distance.

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for R in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

Answer 15

(b)

Expert Solution

To determine

Kinetic energy of the block.

Answer to Problem 45P

The kinetic energy is 0.588 J_.

Explanation of Solution

Write the law of conservation of energy.

KA+UA=KB+UB (II)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KB is the kinetic energy at point B and UB is the potential energy at point B.

At point A, kinetic energy is zero.

At point B, potential energy is zero.

Therefore,

KB=UA (III)

Substitute (I) in (III)

KB=mgR (IV)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for h in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Kinetic energy of the block.

Answer 20

Answer to Problem 45P

The kinetic energy is 0.588 J_.

Answer 21

Explanation of Solution

Write the law of conservation of energy.

KA+UA=KB+UB (II)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KB is the kinetic energy at point B and UB is the potential energy at point B.

At point A, kinetic energy is zero.

At point B, potential energy is zero.

Therefore,

KB=UA (III)

Substitute (I) in (III)

KB=mgR (IV)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 30 cm for h in (I).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(30 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(30×10−2 m)=0.588 J

The gravitational potential energy is 0.588 J_.

Answer 22

(c)

Expert Solution

To determine

Velocity at point B.

Answer to Problem 45P

Velocity is 2.42 m/s_

Explanation of Solution

Write down the equation for kinetic energy.

KB=12mvB2 (V)

Here KB is the kinetic energy at point B, m is the mass of the block, vB is the velocity

Rewrite (V) in terms of vB.

vB=2KBm (VI)

Conclusion:

Substitute 0.588 J for KB and 200 g for m in (VI).

vB=2(0.588 J)(200 g(10−3 kg1 g))=2(0.588 J)(200×10−3 kg)=2.42 m/s

Velocity is 2.42 m/s_

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Velocity at point B.

Answer 27

Answer to Problem 45P

Velocity is 2.42 m/s_

Answer 28

Explanation of Solution

Write down the equation for kinetic energy.

KB=12mvB2 (V)

Here KB is the kinetic energy at point B, m is the mass of the block, vB is the velocity

Rewrite (V) in terms of vB.

vB=2KBm (VI)

Conclusion:

Substitute 0.588 J for KB and 200 g for m in (VI).

vB=2(0.588 J)(200 g(10−3 kg1 g))=2(0.588 J)(200×10−3 kg)=2.42 m/s

Velocity is 2.42 m/s_

Answer 29

(d)

Expert Solution

To determine

Kinetic energy and potential energy at point C.

Answer to Problem 45P

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

Explanation of Solution

Write down the equation for gravitational potential energy

UC=mghC (VII)

Here UC is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and hC is the height at point C.

Write the law of conservation of energy.

KA+UA=KC+UC (VIII)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KC is the kinetic energy at point C and UC is the potential energy at point C.

Rewrite (VIII)

KC=KA+UA−UC

At point A, kinetic energy is zero.

Therefore,

KC=UA−UC (IX)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 20 cm for hc in (VII).

U=(200 g(10−3 kg1 g))(9.8 m/s2)(20 cm(10−2 m1 cm))=(200×10−3 kg)(9.8 m/s2)(20×10−2 m)=0.392 J

Substitute 0.588 J for UA and 0.392 J for UC in (IX)

KC=0.588 J−0.392 J=0.196 J

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

Kinetic energy and potential energy at point C.

Answer 34

Answer to Problem 45P

Potential energy is 0.392 J_.

Kinetic energy is 0.196 J_

Answer 35

Explanation of Solution

Write down the equation for gravitational potential energy

UC=mghC (VII)

Here UC is the gravitational potential energy at point B, m is the mass, g is the acceleration due to gravity and hC is the height at point C.

Write the law of conservation of energy.

KA+UA=KC+UC (VIII)

Here KA is the kinetic energy at point A, UA is the potential energy at point A, KC is the kinetic energy at point C and UC is the potential energy at point C.

Rewrite (VIII)

KC=KA+UA−UC

At point A, kinetic energy is zero.

Therefore,

KC=UA−UC (IX)

Conclusion:

Substitute 200 g for m, 9.8 m/s2 for g and 20 cm for hc in (VII).