JEE MAIN - Physics (2019 - 9th January Evening Slot - No. 23)
A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds ?
850 J
950 J
875 J
900 J
Explanation
Displacement,
x = 3t2 + 5
$$ \therefore $$ v = $${{dx} \over {dt}} = 6t$$
At t = 0, velocity = 6 $$ \times $$ 0 = 0
at t = 5, velocity = 5 $$ \times $$ 6 = 30 m/s
we know from work energy theorem,
Work (W) = change in kinetic energy ($$\Delta $$K)
= $${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$$
= $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ (30)2 $$-$$ 0
= 900 J
x = 3t2 + 5
$$ \therefore $$ v = $${{dx} \over {dt}} = 6t$$
At t = 0, velocity = 6 $$ \times $$ 0 = 0
at t = 5, velocity = 5 $$ \times $$ 6 = 30 m/s
we know from work energy theorem,
Work (W) = change in kinetic energy ($$\Delta $$K)
= $${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$$
= $${1 \over 2}$$ $$ \times $$ 2 $$ \times $$ (30)2 $$-$$ 0
= 900 J
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