Question

A cart having $4$ wheels is drawn with a constant force $F_0$=$16N$ up an inclined plane, making an angle $\alpha ={30}^0$ with the horizontal. The platform of the cart weighs $W_0=18N$ and each of its uniform wheels (cylindrical) weighs $W=2N$. lf the initial velocity is zero, the linear velocity $v_1$ (in $m/s)$ of the cart when it has travelled a distance of $P=4m$ is :
(Given: The wheels roll without slipping and neglect rolling friction)

• A. $1.4$
• B. $2.8$
• C. $9.8$
• D. $4.2$

Answer 1

The correct option is B $2.8$

The wheel is in plane motion,
kinetic energy $E_k=\frac{1}{2}M{v}_c^2+\frac{1}{2}I{\omega }^2$

As the body is a uniform cylinder,
we have, $I=\frac{M{R}^2}{2}$ where R is the radius of wheel.

Also, $\omega =\frac{v_c}{R}$

$\Rightarrow E_k=\frac{1}{2}M{v}_c^2+\frac{1}{4}M{R}^2\frac{V_c^2}{R^2}=\frac{3}{4}M{v}_c^2$

$(E_k{)}_0=0$

$(E_k{)}_1=(E_k{)}_{platform}+4(E_k{)}_{wheel}$

The cart is in translatory motion

$(E_k{)}_1=\frac{1}{2}\left(\frac{W_0}{g}{v}_1^2\right)+4\left(\frac{3}{4}\frac{w}{g}{v}_1^2\right)$ $=\frac{1}{2g}(W_0+6w)v_1^2$

$W_1=$ work done by $F_0=F_{0}l$

$W_2=$ work done by $W_0+4w=-(W_0+4w)\left(l\sin \alpha \right)$
By Work-Energy Theorem,

$(E_k{)}_1=W_1+W_2$

$\Rightarrow \frac{1}{2g}(W_0+6w)v_1^2=[F_0-(W_0+4w\left)sin\alpha \right]l$

$\Rightarrow v_1=\sqrt{\frac{2gl[F_0-(W_0+4w\left)\sin \alpha \right]}{W_0+6w}}$

$=\sqrt{\frac{2\times 9.8\times 4[16-(18+4\times 2\left)\sin {30}^0\right]}{18+6\times 2}}$
$=2.8m/s$