Question

A 30 kg wooden box sits on a smooth concrete floor.

Answer 1

1) The force exerted by the box on the floor is the weight of the box and it is 294 N, and it is equal to the normal reaction exerted by the floor on the box

2) The force required to get the box moving is 102.9 N

3) The force required to keep the box moving at a constant velocity is 67.6 N

4) Find the free-body diagram in attachment

5) The final velocity of the box is 11.8 m/s

Explanation:

1)

The force exerted by the box on the floor is equal to the weight of the box (which is the gravitational attraction exerted by the Earth on the box, which pulls the box downward, and therefore the box pushes downward on the floor with exactly this force). Its magnitude is

$W=mg$

where

m is the mass of the box

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

The box in this problem has a mass of

m = 30 kg

so its weight is

$W=(30)(9.8)=294 N$

Now, according to Newton's third law:

"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"

This means that since the box exerts a force of 294 N on the floor (downward), then the floor exerts an equal and opposite force of 294 N (upward) on the box: this force is called normal reaction, and it is indicated with R.

2)

The force required to put the box in motion is equal to the maximum force of static friction, which is:

$F_{max}=\mu_s R$

where

$\mu_s=0.35$ is the coefficient of static friction

R is the normal reaction

We said in part 1) that the normal reaction is equal to the weight of the box,

R = W = 294 N

So the maximum force of static friction is

$F_{max}=(0.35)(294)=102.9 N$

3)

Once the box is set in motion, the force of static friction is replaced by the force of kinetic friction, which is given by

$F_f = \mu_k R$

where

$\mu_k = 0.23$ is the coefficient of kinetic friction

R = 294 N is the normal reaction

When pushing the box horizontally with a force F, the equation of motion for the box is

$F-F_f = ma$

where a is its acceleration. However, we want to move the box at constant velocity, so

a = 0

Therefore, the force applied must be equal to the force of kinetic friction:

$F-F_f = 0 \rightarrow F=F_f = (0.23)(294)=67.6 N$

4)

The free-body diagram can be found in attachment. We have 4 forces acting on the box:

- The weight, W, acting downward

- The normal reaction R, upward, of same magnitude of the weight

- The pushing force F, forward

- The force of kinetic friction, $F_f$, backward and of same magnitude of F

5)

In this case, the equation of motion along the horizontal axis would be

$F-F_f = ma$

where:

F = 102.9 N is the force applied (calculated at point 2)

$F_f = 67.6 N$ is the force of friction

m = 30 kg is the mass of the box

We can solve the equation to find the acceleration:

$a=\frac{F-F_f}{m}=\frac{102.9-67.6}{30}=1.18 m/s^2$

This acceleration is applied for 10 seconds, so we can find the final velocity with the equation:

v = u + at

where

u = 0 is the initial velocity

$a=1.18 m/s^2$ is the acceleration

t = 10 s is the time

Substituting,

$v=0+(1.18)(10)=11.8 m/s$

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