Chapter 10, Problem 89A

(a)

To determine

The work done by B on the weights by raising it upwards.

Answer to Problem 89A

The work done by the B on the weights is $5.5\times {10}^3\text{J}$ .

Explanation of Solution

Given:

The mass of the weights is $m=240\text{kg}$ .

The weight lifter raised the weights to a distance $d=2.35\text{m}$ .

The weight is lifted at constant speed.

Consider the acceleration due to gravity of the Earth is $g=9.80{\text{m/s}}^2$ .

Formula used:

The expression for the work done by an object is

  $\begin{array}{c}W=F\cdot d\\ W=Fd\cos \theta \end{array}$

Here, $F$ is the force applied, $d$ is the displacement of the object, and $\theta$ is the angle between the force and displacement vector.

Calculation:

Consider the angle between the force vector and the displacement vector is denoted by $\theta$ .

The weight lifter applies a force in the upward direction to lift the weight in the upward direction.

Both the displacement vector and the force vector is in the upward direction.

Hence, the angle between the force and displacement vector is $\theta =0°$ .

The work done by the weight lifter in lifting the weights is,

  $\begin{array}{l}W=Fd\cos \theta \\ W=\left(mg\right)d\cos \theta \\ W=\left(240\text{kg}\times \text{9}\text{.80}{\text{m/s}}^2\right)\times \left(2.35\text{m}\right)\cos \left(0°\right)\\ W=\left(2352\text{kg}\cdot {\text{m/s}}^2\times \frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\times \left(2.35\text{m}\right)\\ W=5.5\times {10}^3\text{J}\end{array}$

Conclusion:

Therefore, the work done by the B on the weights is $5.5\times {10}^3\text{J}$ .

(b)

To determine

The work done by B on the weights by holding it above his head.

Answer to Problem 89A

No work is done by the B on the weights.

Explanation of Solution

Given:

The mass of the weights is $m=240\text{kg}$ .

Consider the acceleration due to gravity of the Earth is $g=9.80{\text{m/s}}^2$ .

Formula used:

The expression for the work done by an object is

  $\begin{array}{c}W=F\cdot d\\ W=Fd\cos \theta \end{array}$

Here, $F$ is the force applied, $d$ is the displacement of the object, and $\theta$ is the angle between the force and displacement vector.

Calculation:

The weight lifter is holding the weight over his head. Hence, the displacement of the weight is $d=0\text{m}$ .

Find the work done by the weight lifter in holding the weights on his head as follows:

  $\begin{array}{l}W=Fd\cos \theta \\ W=\left(mg\right)d\cos \theta \\ W=\left(mg\right)\left(0\right)\cos \theta \\ W=0\text{J}\end{array}$

Conclusion:

Therefore, no work is done by the B on the weights.

(c)

To determine

The work done by B on the weights by lowering it downwards.

Answer to Problem 89A

The work done by the B on the weights is $-5.5\times {10}^3\text{J}$ .

Explanation of Solution

Given:

The mass of the weights is $m=240\text{kg}$ .

The weight lifter lowered the weights to a distance $d=2.35\text{m}$ .

The weight is lifted at constant speed.

Consider the acceleration due to gravity of the Earth is $g=9.80{\text{m/s}}^2$ .

Formula used:

The expression for the work done by an object is

  $\begin{array}{c}W=F\cdot d\\ W=Fd\cos \theta \end{array}$

Here, $F$ is the force applied, $d$ is the displacement of the object, and $\theta$ is the angle between the force and displacement vector.

Calculation:

Consider the angle between the force vector and the displacement vector is denoted by $\theta$ .

While lowering the weights, the angle between the force and displacement vector becomes $\theta =180°$ as these two vectors become anti parallel.

Find the work done by the weight lifter in lowering the weights as follows:

  $\begin{array}{l}W=Fd\cos \theta \\ W=\left(mg\right)d\cos \theta \\ W=\left(240\text{kg}\times \text{9}\text{.80}{\text{m/s}}^2\right)\times \left(2.35\text{m}\right)\cos \left(180°\right)\\ W=-\left(2352\text{kg}\cdot {\text{m/s}}^2\times \frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\times \left(2.35\text{m}\right)\\ W=-5.5\times {10}^3\text{J}\end{array}$

Conclusion:

Therefore, the work done by the B on the weights is $-5.5\times {10}^3\text{J}$ .

(d)

To determine

To identify: Whether some work will be by B on the weights by letting it go and fall back to the ground or not.

Answer to Problem 89A

No work done by the B on the weights.

Explanation of Solution

Given:

The mass of the weights is $m=240\text{kg}$ .

The weight lifter lowered the weights to a distance $d=2.35\text{m}$ .

The weight is lifted at constant speed.

Consider the acceleration due to gravity of the Earth is $g=9.80{\text{m/s}}^2$ .

Formula used:

The expression for the work done by an object is

  $\begin{array}{c}W=F\cdot d\\ W=Fd\cos \theta \end{array}$

Here, $F$ is the force applied, $d$ is the displacement of the object, and $\theta$ is the angle between the force and displacement vector.

Calculation:

The weight lifter let the weight go off and fall back on the ground. Hence, the force applied by the weight lifter on the weights is $F=0\text{N}$ .

Find the work done by the weight lifter in letting the weight go off and fall back to the ground as follows:

  $\begin{array}{l}W=Fd\cos \theta \\ W=\left(0\right)d\cos \theta \\ W=0\text{J}\end{array}$

Hence, no work is done by the B on the weights.

Conclusion:

Therefore, no work is done by the B on the weights.