Question

A ball of mass 50 gram and relative density 0.5 strike the surface of water with velocity 20 metre per second it comes to rest at depth of 2m find the work done by the resting force in water​

Answer 1

Answer:

The answer is 11 J.

Explanation:

Mass of ball = 50 mg = 50/1000 == 0.05 Kg

Density = 0.5  

Velocity "V" = 20 ms-1

Thus  

E1 = 1/2 mv^2 + 0

E2  0 - MG(2)

E1 - Wf = E2  

Where "Wf" is the work done by resting force.

Hence  

Wf = E1 - E2

Wf = 1/2 mv^2 - (-2mg)

Wf = 1/2 mv^2 + 2mg

   = 1/2 x 0.05 x(20)^2  + 2 x 0.05 x 10

   = 11 J