Question

An intern pushes a 97 kg computer cart down the hall. The cart starts from rest and travels 27.634 m in 12.33 s. What force does he apply to the cart?

Answer 1

35.26 N

Step-by-step explanation:

A 97 kg cart is pushed down the hall. The cart starts from rest. The cart travels 27.634 m in 12.33 sec. Since not much is mentioned about the pushing force, it is assumed constant.

So, the cart is under a uniformly accelerated linear motion.

Also, since nothing is mentioned about friction, it is assumed that the surface is smooth.

Consider the equation of motion,
`S=ut+(1)/(2)at^(2)`

Here
`S` is the distance covered,
`u` is the initial velocity,
`a` is the acceleration,
`t` is the time.

On substituting the values,

`27.634=(0)(12.33)+(1)/(2)a(12.33)^(2)\\a=0.364 (m)/(s^(2))`

So, force applied =
`m* a=97*0.364=35.26\text{ }N`

∴ Force applied on the cart = 35.26 N