Question

If a box is pushed across a floor with a force of 130N. The frictional force acting between the box and the floor is 30N, over a time period of 5 sec,

Determine :
1) What is the net force acting on it?
2) Acceleration acting upon the box if it starts from rest and attains a certain Kinetic energy after being pushed 25m?
3) Also find the mass of the box.
4) What will be the kinetic energy and final velocity?

Answer 1

The force acting in the front direction is the 130N.
The frictional force is acting backwards 30N.

1) The net force is 130N - 30N = 100N

2) s = ut + (1/2)at^2 u = 0, Start from rest, s = 25m t =5.

25 = 0*5 + (1/2)* a * 5^2.

25 = 0 + 25/2 * a.

25 = (25/2)a. Divide 25 from both sides.

1 = (1/2)* a. Cross multiply.

2 = a.

a = 2 m/s^2.

3) Mass of the box
Net Force, F = ma
100 = m*2. Divide both sides by 2.

100/2 = m
50 = m.
m = 50 kg.

4) Final velocity , v = u + at.
v = 0 + 2*5 = 10 m/s.

Kinetic Energy, K = (1/2) * mv^2.
= 1/2 * 50 * 10 * 10.
= 2500 J.