Practice portal › Work, Energy and Power › Work-Energy Theorem
Asked in NEET 1989 · Work against resistance
Given: m=10 g=0.01 kg, launch speed 1000 m s⁻¹, landing speed 500 m s⁻¹ at the same level.
Idea: same level means gravity's net work is zero, so all the missing kinetic energy went to the air.
W=1/2m(u²-v²).
=1/2(0.01)(1000²-500²).
=0.005×(1000000-250000)=0.005×750000.
=3750 J.
Reading this solution needs no sign-in. You only sign in to practise against the clock and keep your progress.
Practise this with a timer