Question

. An object of mass 10 kg is dropped from a tower of height 150 m. The body strikes the ground with velocity

40 ms–1. Calculate work done by air friction. [Take g = 10 m/s2

]​

Answer 1

answer: -7000J

Work done by air friction= M.E.(final) - M.E.(initial)

=(1/2×10×(40) ^2 + 0)J - (0 + 10×10×150)J

=8000J - 15000J

= -7000J

(here, -ve sign means work done against motion of object)

Answer 2

Given : An object of mass 10 kg is dropped from a tower of height 150 m

The body strikes the ground with velocity 40 ms–1.

To Find :  work done by air friction

Solution:

object of mass 10 kg is dropped from a tower of height 150 m

h = 150 m

v = 0

PE = mgh = 10 * 10 * 150 = 15000  J

KE = (1/2)mv² = 0  J

TE = PE + KE = 15000 + 0 = 15000 J

The body strikes the ground with velocity 40 ms–1.

v = 40 m/s

h = 0

PE = mgh =  0  J

KE = (1/2)mv² = (1/2)10(40)² =   8000J

TE = PE + KE = 0 + 8000   = 8000 J

work done by air friction  =   Change in Energy

= 8000  - 15000

= -7000 J

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