Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781337026345
Author: Katz
Publisher: Cengage
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Chapter 9, Problem 8PQ

A 537-kg trailer is hitched to a truck. Find the work done by the truck on the trailer in each of the following cases. Assume rolling friction is negligible. a. The trailer is pulled at constant speed along a level road for 2.30 km. b. The trailer is accelerated from rest to a speed of 88.8 km/h. c. The trailer is pulled at constant speed along a road inclined at 12.5° for 2.30 km.

(a)

Expert Solution
Check Mark
To determine

The work done for by the truck on the trailer when the trailer is pulled at constant speed.

Answer to Problem 8PQ

The work done for by the truck on the trailer when the trailer is pulled at constant speed is 0_.

Explanation of Solution

Write the equation of work done.

  Wtot=KfKi                                                                                                    (I)

Here, Wtot is the work done, Kf is the final kinetic energy and Ki is the initial kinetic energy.

At constant speed, the initial and final kinetic energy is same.

  Kf=Ki                                                                                                              (II)

Conclusion:

Substitute, Kf for Ki in equation (I) to find Wtot.

  Wtot=KfKi=0

Thus, the work done for by the truck on the trailer when the trailer is pulled at constant speed is 0_.

(b)

Expert Solution
Check Mark
To determine

The work done for by the truck on the trailer when the trailer is pulled at 88.8km/h.

Answer to Problem 8PQ

The work done for by the truck on the trailer when the trailer is pulled at 88.8km/h is 1.64×105J_.

Explanation of Solution

Write the equation of work done.

  Wtot=KfKi                                                                                                  (III)

Here, Wtot is the work done, Kf is the final kinetic energy and Ki is the initial kinetic energy.

In this case, the initial kinetic energy is zero.

  Ki=0                                                                                                              (IV)

Write the expression for the final kinetic energy.

  Kf=12mv2                                                                                                        (V)

Here, m is the mass and v is the speed.

Rewrite the expression from equation (I).

  Wtot=12mv2                                                                                                    (VI)

Conclusion:

Substitute 537kg for m and 88.8km/h for v in equation (VI) to find Wtot.

  Wtot=12(537kg)[(88.8km/h)(0.277778m/s1km/h)]=(268.5kg)[(88.8km/h)(0.277778m/s1km/h)]=1.64×105J

Thus, the work done for by the truck on the trailer when the trailer is pulled at 88.8km/h is 1.64×105J_.

(c)

Expert Solution
Check Mark
To determine

The work done for by the truck on the trailer when the trailer is pulled at an angle 12.5°.

Answer to Problem 8PQ

The work done for by the truck on the trailer when the trailer is pulled at an angle 12.5° is 2.62×106J_.

Explanation of Solution

Write the equation of work done.

  WT=Wg                                                                                                 (VII)

Here, WT is the work done by the truck and Wg is the work done by gravity.

Write the expression for the work done by gravity.

  Wg=mgrcos(90+θ)                                                                                     (VIII)

Here, m is the mass g is the gravitational acceleration, r is the distance and θ is the angle.

Rewrite the expression from equation (VIII).

  WT=mgrcos(90+θ)                                                                                 (IX)

Conclusion:

Substitute 537kg for m, 9.8m/s2 for g, 2.30km for r and 12.5° for θ in equation (IX) to find WT.

  WT=(537kg)(9.8m/s2)[(2.30km)(1×103m1km)]cos(90°+(12.5°))=(537kg)(9.8m/s2)(2.30×103m)cos(102.5°)=2.62×106J

Thus, the work done for by the truck on the trailer when the trailer is pulled at an angle 12.5° is 2.62×106J_.

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Chapter 9 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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Mechanical work done (GCSE Physics); Author: Dr de Bruin's Classroom;https://www.youtube.com/watch?v=OapgRhYDMvw;License: Standard YouTube License, CC-BY