Question

A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20. (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on it? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 N to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? (e) If the monkey applies a horizontal force of 18.0 N, what is the magnitude of the friction force and what is the box’s acceleration?

Answer 1

Answer:

a) $f=0\ N$

b) $f=6\ N$

c) $F=16\ N$

d) $a=2.45\ m.s^{-2}$

Explanation:

Given:

weight of the box on the horizontal surface, $w=40\ N$

coefficient of static friction between the surface and the box, $\mu_s=0.2$

coefficient of kinetic friction between the surface and the box, $\mu_k=0.2$

a)

When no horizontal force acts on the box then according to the Newton's first law of motion there will be no any force of friction acting on the body but just a vertical component is balanced by the normal reaction.

b)

Now force on the box, $F=6\ N$

So there we have the maximum force of static friction as:

$f_s=\mu_s.N$

here:

N = normal force equal to the weight of the body

$f_s=0.4\times 40$

$f_s=16\ N$

• Now the magnitude of the static frictional force is equal to the applied force on the box. So,

$f=6\ N$

c)

Since we have the maximum static frictional force between the two surfaces as:

$f_s=16\ N$

• So, the applied force must be equal to this limiting value.
• So the applied force must be:

$F=16\ N$

d)

Now when the box has started its motion then the minimum intensity of the force to keep the box moving is equals to the kinetic frictional force:

$F_k=\mu_k.N$

$F_k=0.2\times 40\\F_k=8\ N$

e)

The value of friction force:

Since the box is moving, so the maximum friction is the kinetic friction:

$F_k=8\ N$

The applied force is :

$F=18\ N$

So the acceleration will be due to :

$\Delta F=F-F_k$

$\Delta F=10\ N$

• now we know that:

$a=\frac{\Delta F}{m}$

$a=10\div\frac{40}{9.8}$

$a=2.45\ m.s^{-2}$