Question 140: Work, Energy and Power & Circular Motion

Question

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 ?

Accepted Answer

Displacement,
x = 3t
2
+ 5
$$ 	herefore $$  v = $${{dx} \over {dt}} = 6t$$
At  t = 0,   velocity = 6 $$ 	imes $$ 0 = 0
at  t = 5, velocity = 5 $$ 	imes $$ 6 = 30 m/s
we  know from work energy theorem,
Work (W) = change in kinetic energy ($$\Delta $$K)
= $${1 \over 2}mv_F^2 - {1 \over 2}mv_i^2$$
=  $${1 \over 2}$$ $$ 	imes $$ 2 $$ 	imes $$ (30)
2
$$-$$ 0
= 900 J

Answer

850 J

Answer

950 J

Answer

875 J

Answer

900 J

Work, Energy and Power & Circular Motion

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 ?

Choose the correct answer: