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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
Check Mark
To determine

(a)

The velocity of the block when t=2s .

Answer to Problem 1PP

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

Explanation of Solution

Given:

The mass of the block is m=10kg .

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 Plus Modified Mastering Engineering Revision With Pearson Etext -- Access Card Package (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip  1

Write the formula for Newton’s second law of motion along xaxis .

Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along xaxis and Fx is the resultant of all the forces acting on the block along xaxis .

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 xaxis .

Fx=500N(45)300N=100N

Calculate the acceleration of the block.

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

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

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

Here, ax=ac=10m/s2 .

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

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

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

Expert Solution
Check Mark
To determine

b)

The velocity of the block when t=2s .

Answer to Problem 1PP

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

Explanation of Solution

Given:

The mass of the block is m=10kg .

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 Plus Modified Mastering Engineering Revision With Pearson Etext -- Access Card Package (14th Edition), Chapter 13.4, Problem 1PP , additional homework tip  2

Write the formula for Newton’s second law of motion along xaxis .

Fx=max (I)

Here, m is the mass of the block, ax is the acceleration of the block along xaxis and Fx is the resultant of all the forces acting on the block along xaxis .

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

a=dvdtdv=adt (II)

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

Conclusion:

Refer Figure (1b).

Resolve the forces along xaxis .

Fx=20t= 20t

Calculate the acceleration of the block.

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

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

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

Here, ax=a=2t .

Substitute 2t for a in Equation (II).

dv=2tdt

Integrate at the limits of v=0to v and t=0to2s .

0vdv=022tdt[v]0v=[2t22]02(v0)=(2)2(0)0v=4m/s

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

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Chapter 13 Solutions

Engineering Mechanics: Dynamics Plus Modified Mastering Engineering Revision With Pearson Etext -- Access Card Package (14th Edition)

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