Question

A force of 30N acts on a body of mass 2kg starting from rest up to a distance of 3m.Then the force reduces to 15N and acts in the same directions up to 2m .Calculate the final kinetic energy of the body

Answer 1

Given,
m=2kg
F=30N
Distance covered  (S)= 3m
From F = ma relation
a= F/m
  =30/2
  =15m/s²
From equation of motion
v² = u² + 2aS
v² = (0)² + 2 * 15* 3 (Since motion of body starts from rest)
    = 90
In second case,
the final velocity produced by first force becomes the initial velocity of the next force,such that
v=u
u² = v² = 90
Given ,
Force F= 15N
S = 2m
From F = ma relation
a= F/m
  = 15/2
  = 7.5m/s²
From equation of motion 
v² = u² + 2aS
v² = 90 + 2 * 7.5* 2 (Since u² = v²  )
    = 90 + 30
    = 120
Final kinetic energy = 1/2 * mass of moving body * (final velocity)²
                                =  1/2 * 2 * 120
                                = 120 joules (Ans)

Answer 2

Answer:

120 j

Explanation:

given , F=30N ,m =2kg , s= 3kg

a\q ,

f= ma = a=f \ m = a= 30÷2 = 15.

from eq . of motion ,

(v)2 =(u)2 + 2as

(v)2 = 0 + 2×15×3

(v)2= 90

(v)2=(u)2=90

f=15N , m= 2 kg, s=2

a=15÷3=7.5

(v)2=(u)2 +2as

= 90 + 30

= 120

then ,

ke = 1÷2 m(v)2

= 1÷2× 2×120

= 120 j.