Question

A worker pushes a 50 kg crate a distance of 7.5 m across a level floor. He

Answer 1

a) 73.5 N

b) 551.3 J

c) -551.3 J

d) 0 J

e) 0 J

f) 0 J

g) 0 J

Explanation:

a)

There are two forces acting on the crate:

- The push of the worker, F, in the forward direction

- The frictional force, $F_f=\mu mg$, in the backward direction, where  

$\mu=0.15$ is the coefficient of friction

m = 50 kg is the mass of the crate

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

According to Newton's second law of motion, the net force on the crate must be equal to the product of mass and acceleration, so:

$F-F_f=ma$

However, the crate here is moving with constant velocity, so its acceleration is zero:

$a=0$

So the previous equation becomes:

$F-F_f=0$

And we can find the magnitude of the applied force:

$F=F_f=\mu mg=(0.15)(50)(9.8)=73.5 N$

b)

The work done by the applied force on the crate is

$W_F=Fd cos \theta$

where:

F is the magnitude of the force

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here we have:

F = 73.5 N

d = 7.5 m

$\theta=0^{\circ}$ (the force is applied in the same direction as the displacement)

Therefore,

$W_F=(73.5)(7.5)(cos 0^{\circ})=551.3 J$

c)

The work done by friction on the  crate is:

$W_{F_f}=F_f d cos \theta$

where in this case:

$F_f=73.5 N$ is the magnitude of the force of friction

d = 7.5 m is the displacement of the crate

$\theta=180^{\circ}$, because the displacement is forward and the  force of friction is backward, so they are in opposite direction

Therefore, the work done by the force of friction is:

$W_{F_f}=(73.5)(7.5)(cos 180^{\circ})=-551.3 J$

d)

To find the normal force, we analyze the situation of the force along the vertical direction.

We have two forces on the vertical direction:

- The normal force, N, upward

- The force of gravity, $mg$, downward, where

m = 50 kg is the mass of the crate

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

Since the crate is in equilibrium in this direction, the vertical acceleration is zero, so the two forces balance each other:

$N-mg=0\\N=mg=(50)(9.8)=490 N$

The work done by the normal force is:

$W_N=Nd cos \theta$

In this case, $\theta=90^{\circ}$, since the normal force is perpendicular to the displacement of the crate; therefore, the work done is

$W_N=(490)(7.5)(cos 90^{\circ})=0$

e)

The work done by the gravitational force is:

$W_g=F_g d cos \theta$

where:

$F_g=mg=(50)(9.8)=490 N$ is the gravitational force

d = 7.5 m is the displacement of the crate

$\theta=90^{\circ}$ is the angle between the direction of the gravitational force (downward) and the displacement (forward)

Therefore, the work done by gravity is

$W_g=(490)(7.5)(cos 90^{\circ})=0 J$

f)

The total work done on the crate can be calculated by adding the work done by each force:

$W=W_F+W_{F_f}+W_N+W_g$

where we have:

$W_F=+551.3 J$ is the work done by the applied force

$W_{F_f}=-551.3 J$ is the work done by the frictional force

$W_N=0$ is the work done by the normal force

$W_g=0$ is the work done by the force of gravity

Substituting,

$W=+551.3+(-551.3)+0+0=0 J$

So, the total work is 0 J.

g)

According to the work-energy theorem, the change in kinetic energy of the crate is equal to the work done on it, therefore:

$W=\Delta E_K$

where

W is the work done on the crate

$\Delta E_K$ is the change in kinetic energy of the crate

In this problem, we have:

$W=0$ (total work done on the crate is zero)

Therefore, the change in kinetic energy of the crate is:

$\Delta E_K = W = 0$