Question

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 seconds is:

A.4.50 J
B.7.50 J
C.5.06 J
D.14.06 J

Answer 1

Hint: In order to solve this question, carefully observe the given diagram the area under the F−t graph gives the change in momentum of the block. Then we can calculate the value of kinetic energy by using the relation between the kinetic energy and momentum.

Complete step by step answer:
Area under the F−t graph gives the change in momentum of the block.

Area A = Area of triangle ABO - Area of triangle DCO

Area= $\dfrac{1}{2}(base) (height)$
\begin{align*}
\therefore A &=\dfrac{1}{2}(4)\times (3) −\dfrac{1}{2}(2)\times (1.5)\\
 &=4.5 Ns
\end{align*}
The Initial momentum of the block Pi​=0
Using
\begin{align*}
​A &= {p_{f - }}{p_i}\\
&=4.5-0\\
\Rightarrow &{p_f}​=4.5 Ns\\
\end{align*}
Thus final kinetic energy of the block KE​=$\dfrac{{{p_f}^2}}{{2m}}$
\begin{align*}
KE &​= \dfrac{(4.5)^2}{2(2)}\\
&=5.06J
\end{align*}
Hence the correct option is C

Note:
Alternatively we can use the impulse relation

Given:
At t = 4.5 sec, $\mathop F\limits^ \to = - 2N$
Total Impulse is given by
$\begin{array}{l}
 = \left[ {\dfrac{1}{2} \times 3 \times 4} \right] - \left[ {\dfrac{1}{2} \times 2 \times 1.5} \right]\\
 \Rightarrow I = 6 - 1.5 = 4.5\,\,{\rm{SI}}\,{\rm{Unit}}
\end{array}$
Impulse = change in momentum
\begin{align*}
45 &= 2[v - 0]\\
v &= \dfrac{{45}}{2} = 2.25\,m{s^{ - 1}}\\
K.E &= \dfrac{1}{2} \times 2 \times {\left( {2.25} \right)^2}\\
 &= 5.06\,J
\end{align*}