A force acts on a 2 kg object so that its position is given as a function of time as $\mathrm{x}=3 \mathrm{t}^{2}+ 5$. What is the work done by this force in first 5 seconds?

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Work, Power & Energy

A force acts on a 2 kg object so that its position is given as a function of time as

x = 3 t^{2} + 5

. What is the work done by this force in first 5 seconds?

(A)850 J
(B)950 J
(C)875 J
(D)900 J✓ Correct

Answer: (D)

Solution

Read the motion off

x = 3 t^{2} + 5

: the velocity is

v = \frac{d x}{d t} = 6 t

, so the object starts from rest (

v = 0

t = 0

) and reaches

v = 6 \times 5 = 30 m/s

t = 5 s

. By the work--energy theorem the work done equals the change in kinetic energy,

W = \frac{1}{2} m (v^{2} - v_{0}^{2}) = \frac{1}{2} (2) (3 0^{2} - 0) = 900 J

JEE Main · 2019 · 09 Jan 2019 · Shift 2 · Single correct option

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