Question

a body of mass m falls through height h starting from ground hits the ground with a velocity of (1.2gh)^1/2 the magnitude of work done by air resistance

Answer 1

a body of mass m falls through height h starting from ground hits the ground with a velocity of (1.2gh)^1/2 the magnitude of work done by air resistance

potential energy at height h = mgh

kinetic energy at height h = 0 as initial velocity= 0

total energy = mgh

potential energy at ground = 0 as height = 0

kinetic energy = (1/2)m v^2

= (1/2)m (1.2gh)

= 0.6mgh

work done = energy lost = mgh - 0.6mgh

= 0.4 mgh

Answer 2

answer : workdone by air resistance = -0.28mgh

little mistake is happened. question is ----> a body of mass m falls through height h starting from ground hits the ground with a velocity of 1.2√(gh) the magnitude of work done by air resistance.

a body of mass m falls through height h starting from ground hits the ground with a velocity of 1.2√(gh)

according to work-energy theorem,

change in kinetic energy + change in potential energy = workdone by air resistance

change in kinetic energy = 1/2mv² - 1/2mu²

here, v = 1.2√(gh) and u = 0

so, change in kinetic energy = 1/2m × 1.44gh - 1/2 m × 0² =0.72mgh

change in potential energy = final potential energy - initial potential energy

= 0 - mgh = -mgh

so, workdone by air resistance = 0.72mgh - mgh

= -0.28mgh