Question

1. A student lifts a box of books that weighs 185 N. The box is

Answer 1

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

$W = \Delta U = (mg) \Delta h$

where

mg is the weight of the object

$\Delta h$ is the change in height

For the box in this problem,

mg = 185 N

$\Delta h = 0.800 m$

Substituting into the equation, we find:

$W=(185)(0.800)=148 J$

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

$W=Fd$

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

$W=(825)(35)=28875 J$

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

$W=Fd$

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

$W' = 2(28875)=57750 J$

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

$F = mg$

where

m = 0.180 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity

Solving the equation,

$F=(0.180)(9.8)=1.76 N$

Now we find the work done by gravity using the same formula applied before:

$W=Fd$

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

$W=(1.76)(2.5)=4.4 J$

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

$\Delta h = 1.2 m$

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

$W=mg \Delta h$

Solving for m, we find

$m=\frac{W}{g \Delta h}$

And substituting the numerical values, we find the mass of the box:

$m=\frac{7000}{(9.8)(1.2)}=595.2 kg$

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

$W=mg \Delta h$

Where $\Delta h$ is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of $\Delta h$ is the same for both of them, so the work they have done is exactly the same.