Question

A block of mass $8\text{kg}$, moving in x-direction with a constant speed of $15\text{m/s}$, is subjected to a retarding force of $F=10x\text{J/m}$ during its travel from $x=10\text{m}$ to $x=15\text{m}$. Find its final kinetic energy.

• A. $300\text{J}$
• B. $325\text{J}$
• C. $275\text{J}$
• D. $350\text{J}$

Answer 1

The correct option is C $275\text{J}$

Given, Mass of block, $\left(M\right)=8\text{kg}$
Initial Speed, $\left(v_i\right)=15\text{m/s}$
Force $\left(F\right)=-10x\text{J/m}$
Initial position $\left(x_1\right)=10\text{m}$
Final position $\left(x_2\right)=15\text{m}$
We know, work energy theorem
$W_{\text{All-forces}}=\Delta K.E$
Using, Workdone by variable force and work-energy theorem
$W=\int Fdx$
$W=K.E_f-K.E_i$
We get,
$\underset{10}{\overset{15}{\int }}-10xdx=K.E_f-\frac{1}{2}\times 8\times {15}^2$
$\Rightarrow \frac{-10}{2}[x^2{]}_{10}^{15}=K.E_f-900$
$\Rightarrow -5\times (225-100)+900=K.E_f$
$\Rightarrow K.E_f=275\text{J}$
Final kinetic energy of block will be $275\text{J}$.
Hence, option(c) is the correct answer.