Question

3. Yumyum pulls a box to the left with 300 N of force. Mamu also pulls the box to the left with 250 N of force. The box

has a mass of 25 kg.
a. What is the net force acting on the box?
b. What is the box's acceleration?

Answer 1

The net force acting on the box is 550 N (going leftwards) and the box's acceleration is 22 meters per second squared.

Step-by-step explanation:

A)

The net force acting on the box is the sum of the forces. Since Yumyum pulls with a force of 300 N and Mamu pulls with a force of 250 N, both to the left, the net force (in the x-direction) will be:

`\displaystyle \sum F _x = (300 \text{ N}) + (250 \text{ N}) = 550 \text{ N}`

B)

The net force is equal to the object's mass multiplied by its acceleration. That is:

`\displaystyle \sum F_x = ma`

Substitute known values:

`\displaystyle (550 \text{ N}) = (25 \text{ kg})a`

Find a:

`\displaystyle a = \frac{550 \text{ N}}{25 \text{ kg}} = 22\text{ m/s$^2$}`

In conclusion, the net force acting on the box is 550 N (going leftwards) and the box's acceleration is 22 meters per second squared.