Chapter 6, Problem 42QAP

To determine

(a)

Work done by the assistant for the football team in carrying a 30.0 kg cooler from the top row of the stadium to the bench area located 20.0 m below.

Answer to Problem 42QAP

The work done by the assistant for the football team in carrying a 30.0 kg cooler from the top row of the stadium to the bench area located 20.0 m below is $-5.88\times {10}^3\text{ J}$.

Explanation of Solution

Given:

Mass of the cooler

  $m=30.0\text{ kg}$

Vertical displacement of the cooler

  $\overrightarrow{h}=20.0\text{ m}\left(-\widehat{j}\right)$

Acceleration due to gravity

  $\overrightarrow{g}=9.80{\text{ m/s}}^2\left(-\widehat{j}\right)$

Formula used:

The work done W by a force $\overrightarrow{F}$ is equal to the scalar (dot) product of the force and the displacement $\overrightarrow{d}$ of the point of application of the force.

  $W=\overrightarrow{F}·\overrightarrow{d}$

Calculation:

The weight $F_G$ ( gravitational force) of the cooler acts downwards, while the force F exerted by the assistant acts in the upward direction. If the cooler moves with constant velocity, the force F is equal in magnitude to the weight $F_G$ but is in a direction opposite to it, keeping the net force on the cooler to have a value zero.

  $\overrightarrow{F}=-{\overrightarrow{F}}_G$

Write an expression for ${\overrightarrow{F}}_G$. Assume the + x direction along the right and +y direction upwards.

  ${\overrightarrow{F}}_G=m\overrightarrow{g}$

Therefore,

  $\begin{array}{c}\overrightarrow{F}=-{\overrightarrow{F}}_G\\ =-m\overrightarrow{g}\end{array}$

Write an expression for the work done by the force exerted by the assistant.

  $W=\overrightarrow{F}·\overrightarrow{h}=-m\overrightarrow{g}·\overrightarrow{h}$

Calculate the work done by substituting the known values in the above equation.

  $\begin{array}{c}W=-m\overrightarrow{g}·\overrightarrow{h}\\ =-\left(30.0\text{ kg}\right)\left[9.80{\text{ m/s}}^2\left(-\widehat{j}\right)\right]\left[20.0\text{ m}\left(-\widehat{j}\right)\right]\\ =-5.88\times {10}^3\text{ J}\end{array}$

Conclusion:

Thus, the work done by the assistant for the football team in carrying a 30.0 kg cooler from the top row of the stadium to the bench area located 20.0 m below is $-5.88\times {10}^3\text{ J}$.

To determine

(b)

The work done by the gravity on the cooler.

Answer to Problem 42QAP

The work done by the gravity on the cooler is $5.88\times {10}^3\text{ J}$.

Explanation of Solution

Given:

Mass of the cooler

  $m=30.0\text{ kg}$

Vertical displacement of the cooler

  $\overrightarrow{h}=20.0\text{ m}\left(-\widehat{j}\right)$

Acceleration due to gravity

  $\overrightarrow{g}=9.80{\text{ m/s}}^2\left(-\widehat{j}\right)$

Formula used:

The work done W by a force $\overrightarrow{F}$ is equal to the scalar (dot) product of the force and the displacement $\overrightarrow{d}$ of the point of application of the force.

  $W=\overrightarrow{F}·\overrightarrow{d}$

Calculation:

Write an expression for the work done by gravitational force.

  $W_G=m\overrightarrow{g}·\overrightarrow{h}$

Substitute the known values of the variables in the expression.

  $\begin{array}{c}{W}_G=m\overrightarrow{g}·\overrightarrow{h}\\ =\left(30.0\text{ kg}\right)\left[9.80{\text{ m/s}}^2\left(-\widehat{j}\right)\right]\left[15.0\text{ m}\left(-\widehat{j}\right)\right]\\ =5.88\times {10}^3\text{ J}\end{array}$

Since the displacement is in the direction of the force, the work done is positive.

Conclusion:

Thus, the work done by the gravity on the cooler is $5.88\times {10}^3\text{ J}$.