While a $2.0$ kilogram ball falls $50.0$ meters, the force of air resistance does $- 25 J o u l e s$ of work on it. How much does the ball's kinetic energy change during the fall?

980 J o u l e s

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75 J o u l e s

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955 J o u l e s

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- 1005 J o u l e s

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- 980 J o u l e s

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Solution

The correct option is B $955 J o u l e s$
Given : $m = 2.0$ kg $h = 50.0$ m
Work done by gravity $W_{g} = m g h = 2.0 \times 9.8 \times 50.0 = 980$ Joules
Work done by air resistance $W_{f} = - 25$ Joules
Using work-energy theorem : $W_{n e t} = Δ K . E$
$∴$ $Δ K . E = W_{g} + W_{f} = 980 - 25 = 955$ Joules

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While a 2.0 kilogram ball falls 50.0 meters, the force of air resistance does −25Joules of work on it. How much does the ball's kinetic energy change during the fall?

The correct option is B 955JoulesGiven : m=2.0 kg h=50.0 m Work done by gravity Wg=mgh=2.0×9.8×50.0=980 JoulesWork done by air resistance Wf=−25 JoulesUsing work-energy theorem : Wnet=ΔK.E∴ ΔK.E=Wg+Wf=980−25=955 Joules

While a $2.0$ kilogram ball falls $50.0$ meters, the force of air resistance does $- 25 J o u l e s$ of work on it. How much does the ball's kinetic energy change during the fall?