Work Energy and Power - Online Test

angle betwwen force and displacement is 90∘

W=Fscos90∘=0

95% potential energy is converted in kinetic energy.

applying conservation of mechanical energy between horizontal and lowermost points

mgl×95100=12mv2gl×95100=12v2v=2×gl×95100−−−−−−−−−−√=2×9.8×1.5×95100−−−−−−−−−−−−−−−√=5.3m/s

from conservation of linear momentum initial momentum = final momentum

p⃗ i=p⃗ f(500×0)+[500×(−2)]=(500+500)v−1000=1000vv=−1m/s

negative sign indicate that cars moves to the left.

Mass of the body, m= 0.5 kg

Velocity of the body  v=ax3/2

a=5m−1/2s−1

Initial velocity at x = 0 is u = 0

Final velocity at x = 2 m is   v=102–√m/s

work done = Change in kinetic energy

W=Kf−KiW=12mv2−0W=12×0.5×(102–√)2=12×0.5×200=50J

Work done by force applied against gravity for one lift will be

W=Fs=mgh=10×9.8×0.5=49J

So work done for 1000 lifts = 49 x 1000 = 49000 J

Work done by force applied against gravity for one lift will be

W=Fs=mgh=10×9.8×0.5=49J

So work done for 1000 lifts = 49 x 1000 = 49000 J

efficiency = Work done / energy from fat

energy from fat used =Work done / efficiency

E=Wη=490000.2=245000J

Energy from per kilogram fat =  3.8 ×$\times$ 107 J

Fat used =2450003.8×107=64473×10−7=6.45×10−3Kg