A horizontal force of 90N is used to push a box across a horizontal floor.?

Question

The frictional force of the box is 50N.
What is the gain in kinetic energy of the box when it is moved through a distance of 6.0m?
1- 240J
2- 300J
3- 540J
4- 840J

Answer 1

Ke =0.5mV^2 or
Ke= Work done with froce applied - Work done against friction
Ke= FaS - f S
Ke= (Fa - f )S
Ke= (90 - 50) 6.0
Ke= 40 x 6.0= 240J

Answer 2

First find the net force experienced by the box:

F = 90 - 50 = 40 N

Now the work done by the net force is equal to the change in kinetic energy of the box so:

change in KE = Work = Net Force * distant

Change in KE = 50 N* 6 m = 300 J

Answer is Number 2.

Answer 3

F=ma
Driving force - Friction = ma
90 - 50 = ma
ma= 40 N
a=40/m

V²=2as+U²
V²-U²=2(40/m)(6)
V²=480/m

Gain in K.E = 1/2 x mass x (V²-U²)
Gain in K.E = 1/2 x m x (480/m)
Gain in K.E = 1/2 x 480
Gain in K.E = 240 J

1- 240J


Very simple way :-
Workdone = Gain in K.E
Work done = Force x distance moved in direction of force
Force = 90 -50 = 40 N
Distance = 6.0 m
Workdone = 40 x 6.0 = 240 J
Gain in K.E = 240 J