In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Question

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Answer 1

Textbook Question

Chapter 13.4, Problem 1PP

In each case, determine its velocity when t = 2 s if v = 0 when t = 0. Chapter 13.4, Problem 1PP, In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Expert Solution

To determine

(a)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 20 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for velocity of the block at constant acceleration as a function of time.

v=v0+act (II)

Here, v is the final velocity, v0 is the initial velocity, ac is the constant acceleration and t is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along x−axis .

∑Fx=500 N(45)−300 N=100 N

Calculate the acceleration of the block.

Substitute 100 N for ∑Fx , 10 kg for m in Equation (I).

100 N=(10 kg)axax=10010ax=10 m/s2

Calculate the velocity of the block at the time t= 2 s .

Here, ax=ac=10 m/s2 .

Substitute 0 for v0 , 10 m/s2 for ac and 2 s for t in Equation (II).

v=0+(10 m/s2)(2 s)=0+20=20 m/s

Thus, the velocity of the block at the time t=2 s is 20 m/s .

Expert Solution

To determine

b)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 4 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for acceleration (a) of the block.

a=dvdtdv=a dt (II)

Here, dvdt is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along x−axis .

∑Fx=20t= 20t

Calculate the acceleration of the block.

Substitute 20t for ∑Fx , 10 kg for m in Equation (I).

20t=(10 kg)axax=20t10ax=2t

Calculate the velocity of the block at the time t= 2 s .

Here, ax=a=2t .

Substitute 2t for a in Equation (II).

dv=2t dt

Integrate at the limits of v=0 to v and t=0 to 2 s .

∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s

Thus, the velocity of the block at the time t=2 s is 4 m/s .

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Answer 2

Textbook Question

Chapter 13.4, Problem 1PP

In each case, determine its velocity when t = 2 s if v = 0 when t = 0. Chapter 13.4, Problem 1PP, In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Expert Solution

To determine

(a)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 20 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for velocity of the block at constant acceleration as a function of time.

v=v0+act (II)

Here, v is the final velocity, v0 is the initial velocity, ac is the constant acceleration and t is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along x−axis .

∑Fx=500 N(45)−300 N=100 N

Calculate the acceleration of the block.

Substitute 100 N for ∑Fx , 10 kg for m in Equation (I).

100 N=(10 kg)axax=10010ax=10 m/s2

Calculate the velocity of the block at the time t= 2 s .

Here, ax=ac=10 m/s2 .

Substitute 0 for v0 , 10 m/s2 for ac and 2 s for t in Equation (II).

v=0+(10 m/s2)(2 s)=0+20=20 m/s

Thus, the velocity of the block at the time t=2 s is 20 m/s .

Expert Solution

To determine

b)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 4 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for acceleration (a) of the block.

a=dvdtdv=a dt (II)

Here, dvdt is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along x−axis .

∑Fx=20t= 20t

Calculate the acceleration of the block.

Substitute 20t for ∑Fx , 10 kg for m in Equation (I).

20t=(10 kg)axax=20t10ax=2t

Calculate the velocity of the block at the time t= 2 s .

Here, ax=a=2t .

Substitute 2t for a in Equation (II).

dv=2t dt

Integrate at the limits of v=0 to v and t=0 to 2 s .

∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s

Thus, the velocity of the block at the time t=2 s is 4 m/s .

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Answer 3

Textbook Question

Chapter 13.4, Problem 1PP

In each case, determine its velocity when t = 2 s if v = 0 when t = 0. Chapter 13.4, Problem 1PP, In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Expert Solution

To determine

(a)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 20 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for velocity of the block at constant acceleration as a function of time.

v=v0+act (II)

Here, v is the final velocity, v0 is the initial velocity, ac is the constant acceleration and t is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along x−axis .

∑Fx=500 N(45)−300 N=100 N

Calculate the acceleration of the block.

Substitute 100 N for ∑Fx , 10 kg for m in Equation (I).

100 N=(10 kg)axax=10010ax=10 m/s2

Calculate the velocity of the block at the time t= 2 s .

Here, ax=ac=10 m/s2 .

Substitute 0 for v0 , 10 m/s2 for ac and 2 s for t in Equation (II).

v=0+(10 m/s2)(2 s)=0+20=20 m/s

Thus, the velocity of the block at the time t=2 s is 20 m/s .

Expert Solution

To determine

b)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 4 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for acceleration (a) of the block.

a=dvdtdv=a dt (II)

Here, dvdt is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along x−axis .

∑Fx=20t= 20t

Calculate the acceleration of the block.

Substitute 20t for ∑Fx , 10 kg for m in Equation (I).

20t=(10 kg)axax=20t10ax=2t

Calculate the velocity of the block at the time t= 2 s .

Here, ax=a=2t .

Substitute 2t for a in Equation (II).

dv=2t dt

Integrate at the limits of v=0 to v and t=0 to 2 s .

∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s

Thus, the velocity of the block at the time t=2 s is 4 m/s .

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Answer 4

Textbook Question

Chapter 13.4, Problem 1PP

In each case, determine its velocity when t = 2 s if v = 0 when t = 0. Chapter 13.4, Problem 1PP, In each case, determine its velocity when t = 2 s if v = 0 when t = 0.

Expert Solution

To determine

(a)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 20 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for velocity of the block at constant acceleration as a function of time.

v=v0+act (II)

Here, v is the final velocity, v0 is the initial velocity, ac is the constant acceleration and t is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along x−axis .

∑Fx=500 N(45)−300 N=100 N

Calculate the acceleration of the block.

Substitute 100 N for ∑Fx , 10 kg for m in Equation (I).

100 N=(10 kg)axax=10010ax=10 m/s2

Calculate the velocity of the block at the time t= 2 s .

Here, ax=ac=10 m/s2 .

Substitute 0 for v0 , 10 m/s2 for ac and 2 s for t in Equation (II).

v=0+(10 m/s2)(2 s)=0+20=20 m/s

Thus, the velocity of the block at the time t=2 s is 20 m/s .

Expert Solution

To determine

b)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 4 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for acceleration (a) of the block.

a=dvdtdv=a dt (II)

Here, dvdt is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along x−axis .

∑Fx=20t= 20t

Calculate the acceleration of the block.

Substitute 20t for ∑Fx , 10 kg for m in Equation (I).

20t=(10 kg)axax=20t10ax=2t

Calculate the velocity of the block at the time t= 2 s .

Here, ax=a=2t .

Substitute 2t for a in Equation (II).

dv=2t dt

Integrate at the limits of v=0 to v and t=0 to 2 s .

∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s

Thus, the velocity of the block at the time t=2 s is 4 m/s .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 20 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for velocity of the block at constant acceleration as a function of time.

v=v0+act (II)

Here, v is the final velocity, v0 is the initial velocity, ac is the constant acceleration and t is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along x−axis .

∑Fx=500 N(45)−300 N=100 N

Calculate the acceleration of the block.

Substitute 100 N for ∑Fx , 10 kg for m in Equation (I).

100 N=(10 kg)axax=10010ax=10 m/s2

Calculate the velocity of the block at the time t= 2 s .

Here, ax=ac=10 m/s2 .

Substitute 0 for v0 , 10 m/s2 for ac and 2 s for t in Equation (II).

v=0+(10 m/s2)(2 s)=0+20=20 m/s

Thus, the velocity of the block at the time t=2 s is 20 m/s .

Expert Solution

To determine

b)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 4 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for acceleration (a) of the block.

a=dvdtdv=a dt (II)

Here, dvdt is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along x−axis .

∑Fx=20t= 20t

Calculate the acceleration of the block.

Substitute 20t for ∑Fx , 10 kg for m in Equation (I).

20t=(10 kg)axax=20t10ax=2t

Calculate the velocity of the block at the time t= 2 s .

Here, ax=a=2t .

Substitute 2t for a in Equation (II).

dv=2t dt

Integrate at the limits of v=0 to v and t=0 to 2 s .

∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s

Thus, the velocity of the block at the time t=2 s is 4 m/s .

Answer 8

Expert Solution

To determine

(a)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 20 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for velocity of the block at constant acceleration as a function of time.

v=v0+act (II)

Here, v is the final velocity, v0 is the initial velocity, ac is the constant acceleration and t is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along x−axis .

∑Fx=500 N(45)−300 N=100 N

Calculate the acceleration of the block.

Substitute 100 N for ∑Fx , 10 kg for m in Equation (I).

100 N=(10 kg)axax=10010ax=10 m/s2

Calculate the velocity of the block at the time t= 2 s .

Here, ax=ac=10 m/s2 .

Substitute 0 for v0 , 10 m/s2 for ac and 2 s for t in Equation (II).

v=0+(10 m/s2)(2 s)=0+20=20 m/s

Thus, the velocity of the block at the time t=2 s is 20 m/s .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a)

The velocity of the block when t=2 s .

Answer 13

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 20 m/s .

Answer 14

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1a).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 1

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for velocity of the block at constant acceleration as a function of time.

v=v0+act (II)

Here, v is the final velocity, v0 is the initial velocity, ac is the constant acceleration and t is the time.

Conclusion:

Refer Figure (1a).

Resolve the forces along x−axis .

∑Fx=500 N(45)−300 N=100 N

Calculate the acceleration of the block.

Substitute 100 N for ∑Fx , 10 kg for m in Equation (I).

100 N=(10 kg)axax=10010ax=10 m/s2

Calculate the velocity of the block at the time t= 2 s .

Here, ax=ac=10 m/s2 .

Substitute 0 for v0 , 10 m/s2 for ac and 2 s for t in Equation (II).

v=0+(10 m/s2)(2 s)=0+20=20 m/s

Thus, the velocity of the block at the time t=2 s is 20 m/s .

Answer 15

Expert Solution

To determine

b)

The velocity of the block when t=2 s .

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 4 m/s .

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .

Write the formula for acceleration (a) of the block.

a=dvdtdv=a dt (II)

Here, dvdt is the rate of change of velocity with respect to time.

Conclusion:

Refer Figure (1b).

Resolve the forces along x−axis .

∑Fx=20t= 20t

Calculate the acceleration of the block.

Substitute 20t for ∑Fx , 10 kg for m in Equation (I).

20t=(10 kg)axax=20t10ax=2t

Calculate the velocity of the block at the time t= 2 s .

Here, ax=a=2t .

Substitute 2t for a in Equation (II).

dv=2t dt

Integrate at the limits of v=0 to v and t=0 to 2 s .

∫0vdv=∫022t dt[v]0v=[2t22]02(v−0)=(2)2−(0)0v=4 m/s

Thus, the velocity of the block at the time t=2 s is 4 m/s .

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

b)

The velocity of the block when t=2 s .

Answer 20

Answer to Problem 1PP

The velocity of the block at the time t=2 s is 4 m/s .

Answer 21

Explanation of Solution

Given:

The mass of the block is m=10 kg .

Initially the velocity of the block is v=0 at t=0 .

The free body diagram of the block as shown in Figure (1b).

Engineering Mechanics: Dynamics (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip 2

Write the formula for Newton’s second law of motion along x−axis .

∑Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along x−axis and ∑Fx is the resultant of all the forces acting on the block along x−axis .