Question

A body of mass 1.0 kg in initially at rest is moved by a horizontal force of 0.5 N on a smooth frictionless table . Calculate the work done by the force in 10s ans show that this is equal to the change in kinetic energy of the body .

Answer 1

Hi

Here is your answer,

Accleration of a body, 
 
                             a = F/m = 0.5/1.0 = 0.5 ms⁻²

Distance travelled, s = ut + 1/2 at²

           → s = 0+1/2 × 0.5(10)² = 25 m

Work done = F × s = 0.5 × 25 = 12.5 J 

           v = u + at = 0 + 0.5 × 10 = 5 ms⁻¹

Change in Kinetic energy = 1/2 m(v² - u²)

          = 1/2 ×  1.0 (5² - 0) = 12.51 



Hope it helps you !

Answer 2

Hi friend,

To prove:-

Work done= 1/2m(v²-u²)

W= F×s

F= ma

a= F/m

a= 0.5/1

a= 0.5m/s²

Distance(S)= ut+1/2 at²

S= 0+1/2×0.5×(10)²

S= 25m.

W= F×s

W= 0.5×25

W= 12.5J.....................(1)

K.E= 1/2 m(v²-u²)

V= at

V= 0.5×10

V=5 m/s.

K.E= 1/2×1×(5)²

K.E= 12.5 J.......(2).

(1)=(2)

Hence it us verified.