Question

A $2\text{kg}$ body (initially at rest) is displaced by a variable force $F=(3x^2+15)\text{N}$ on a rough horizontal plane with coefficient of kinetic friction $\mu =0.5$. Find the speed of the block, after the block has travelled a distance of $5\text{m}$.

• A. $v=5\sqrt{3}\text{m/s}$
• B. $v=\sqrt{3}\text{m/s}$
• C. $v=5\text{m/s}$
• D. $v=5\sqrt{6}\text{m/s}$

Answer 1

The correct option is D $v=5\sqrt{6}\text{m/s}$

Mass of body, $m=2\text{kg}$
Force acting on the body $F=(3x^2+15)\text{N}$
Coefficient of kinetic friction, $\mu =0.5$
Due to sliding on the horizontal surface, kinetic friction will be acting on the body in a direction opposite to its motion.
$f_k=\mu N=\mu mg$

Work done by friction on the body,
$W_f=Fd\cos \theta =\left(\mu mg\right)\times x\times \left(\cos {180}^{\circ }\right)$
$\Rightarrow W_f=-\mu mgx$
$\therefore W_f=-\left(0.5\right)\times 2\times 10\times 5=-50\text{J}$

Work done by variable force $\left(F\right)$ on the body is
$W_F=\int Fdx$
Putting the limits from $x=0\to x=5\text{m}$,
$\Rightarrow W_F={\int }_{x=0}^{x=5}(3x^2+15)dx$
$={[3\times \frac{x^3}{3}]}_0^5+{\left[15x\right]}_0^5$
$=(125-0)+(75-0)=200\text{J}$

Work done by gravity $W_{mg}=0$ (because displacement is horizontal)

Thus, total work done by the external forces on the body is:
$W_{\text{ext}}=W_N+W_{mg}+W_f+W_F$
$\therefore W_{\text{ext}}=0+0-50+200=150\text{J}...\left(i\right)$
Initial kinetic energy, $K{E}_i=0$
Let $v$ be the velocity at $x=5\text{m}$.
Final kinetic energy $K{E}_f=\frac{1}{2}m{v}^2=\frac{1}{2}\times 2\times v^2$
$\therefore \Delta KE=K{E}_f-K{E}_i=v^2....\left(ii\right)$

Applying work energy theorem on the body:
$W_{\text{ext}}=\Delta KE$
From Eq $\left(i\right)\&\left(ii\right)$,
$150=v^2$
$\therefore v=\sqrt{150}=5\sqrt{6}\text{m/s}$