Chapter 10.1, Problem 1PP

a.

To determine

To find: The effect on work done by doubling the force and keeping displacement the same.

Answer to Problem 1PP

The work done gets doubled when the force is doubled and displacement is kept the same.

Explanation of Solution

Given:

A 105-g hockey puck is sliding on ice and a player exerts a constant force of 4.50-N on it so that it moves for a distance of 0.15 m. Now, the force is doubled, so that the new force is 9.0-N. The distance is kept the same.

Formula Used:

When a force F is exerted on an object in some direction and if the material gets displaced by a distance d in the same direction, the work done on it is given by

  $W=Fd$

Calculation:

Here, initially, the force applied on the horse puck by the player is 4.50-N and it gets displaced by a distance 0.150 m. The corresponding work done on the object is given by

  $\begin{array}{c}W=Fd\\ =4.5\times 0.15\\ =0.675\text{ J}\end{array}$

Now, when the force is doubled to 9.0-N keeping the distance the same, the corresponding work done is given by

  $\begin{array}{c}{W}_{new}=F_{new}d\\ =9\times 0.15\\ =1.35\text{ J}\end{array}$

Based on this, the effect on the work done due to the doubling of the force can be expressed as

  $\begin{array}{c}\frac{W_{new}}{W}=\frac{1.35}{0.675}\\ =2\\ W_{new}=2W\end{array}$

Thus, the work done on the puck also gets doubled.

Conclusion:

The work done on the puck gets doubled on doubling the applied force, keeping the distance the same.

b.

To determine

The work done on the puck.

Answer to Problem 1PP

The work done on the puck will be 0.3375 J.

Explanation of Solution

Given:

A 105-g hockey puck is sliding on ice and a player exerts a constant force of 4.50-N on it so that it moves for a distance of 0.15 m. Now, the distance is halved to 0.075 m and the force is kept the same.

Formula Used:

When a force F is exerted on an object in some direction and if the material gets displaced by a distance d in the same direction, the work done on it is given by

  $W=Fd$

Calculation:

Here, initially, the force applied on the horse puck by the player is 4.50-N and it gets displaced by a distance 0.150 m. The corresponding work done on the object is given by

  $\begin{array}{c}W=Fd\\ =4.5\times 0.15\\ =0.675\text{ J}\end{array}$

Now, when the distance is halved by keeping the force the same, the corresponding work done is given by

  $\begin{array}{c}{W}_{new}=F{d}_{new}\\ =4.5\times 0.075\\ =0.3375\text{ J}\end{array}$

Conclusion:

Therefore, the work done on the puck will be 0.3375 J.