Question

4. Johnny exerts a 3.55 N rightward force on a 0.200-kg box to accelerate it across a low-friction track. If the total resistance force to the motion of the cart is 0.52 N, then
a) What is the box's acceleration?
b) What is the gravitational force?
c) What is the normal force?

Answer 1

a) $15.2 m/s^2$

b) 1.96 N

c) 1.96 N

Explanation:

a)

To find the box's acceleration, we have to find first the net force acting on the box in the horizontal direction.

We have:

- The forward force of 3.55 N

- The backward, resistive force of 0.52 N

So, the net force forward is

$\sum F=3.55-0.52=3.03 N$

Now we can find the acceleration by using Newton's second law of motion, which states that:

$\sum F=ma$

where

m = 0.200 kg is the mass of the box

a is its acceleration

And solving for a, we find the acceleration:

$a=\frac{\sum F}{m}=\frac{3.03}{0.200}=15.2 m/s^2$

b)

The gravitational force on an object is the force with which the object is pulled towards the ground by the Earth.

It is given by

$W=mg$

where

m is the mass of the object

g is the gravitational field strength

In this problem we have

m = 0.200 kg is the mass of the box

$g=9.8 m/s^2$ is the gravitational field strength

So, the gravitational force on the box is

$W=(0.200)(9.8)=1.96 N$

c)

The normal force is the reaction force exerted by the floor on the box, in the upward direction.

In order to find the magnitude of this force, we apply Newton's second law of motion along the vertical direction.

We have two forces in this direction:

- The gravitational force, W, downward

- The normal force, N, upward

So the net force is

$\sum F=N-W$

According to Newton's second law,

$\sum F=ma$

However, the box is at rest in the vertical direction, so the vertical acceleration is zero:

$a=0$

This means that the net force is zero:

$\sum F=0$

And so, we can find the normal force:

$N-W=0\\N=W=1.96 N$