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Work, Power & Energy

Section: Physics

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Question : 18 of 22

Marks: +1, -0

A block of mass 2 kg is free to move along the x-axis. It is at rest and from t = 0 onwards, it is subjected to a time-dependent force F(t) in the x- direction. The force F(t) varies with t as shown in the figure. The kinetic energy of the block after 4.5 s is

[JEE Adv 2010 P2]

4.50 J
7.50 J
5.06 J
14.06 J

Solution:

Area under F–t graph is impulse

\vec{J}

=

\Delta\vec{p}

.

Area =

\left(\frac{1}{2} \times 4 \times 3\right) + \frac{1}{2} \times 1.5 \times (-F')

From the figure shown here, we have

\frac{4}{3}

=

\frac{F'}{15}

⇒ F' = 2 N

Therefore, the area is

6 – 1.5 = 4.5 kg-m/s = m(v – u) = 2v (as u = 0)

⇒ v = 2.25 m/s

Therefore,

K.E. =

\frac{1}{2} m v^2

=

\frac{1}{2} \times 2 \times (2.5)^2

= 5.06 J

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