Question

A block of mass $m=1\text{kg}$ is at rest with respect to a rough wedge placed in a lift as shown. The the lift starts moving up from rest with an acceleration of $a=2{\text{m/s}}^2$ and the block remains at rest with respect to wedge. Then in $4\text{seconds}$ of motion (if $\theta ={30}^{\circ }$ and $g=10{\text{m/s}}^2$) work done on block.


$\begin{array}{cccc}& \text{Column-I}& & \text{Column-II}\\ \text{(A)}& \text{By gravity (in magnitude)}& \text{(P)}& 144\text{J}\\ \text{(B)}& \text{By normal reaction}& \text{(Q)}& 32\text{J}\\ \text{(C)}& \text{By frictional force}& \text{(R)}& 160\text{J}\\ \text{(D)}& \text{By all the forces}& \text{(S)}& 48\text{J}\end{array}$
Which of the following is the only corect combination?

• A. $A\to P,B\to R,C\to Q,D\to S$
• B. $A\to Q,B\to S,C\to R,D\to P$
• C. $A\to R,B\to P,C\to S,D\to Q$
• D. $A\to S,B\to Q,C\to P,D\to R$

Answer 1

The correct option is C $A\to R,B\to P,C\to S,D\to Q$

Given,
mass of the block, $m=1\text{kg}$,
acceleration, $a=2{\text{m/s}}^2$,
$\theta ={30}^{\circ }$,
acceleration due to gravity, $g=10{\text{m/s}}^2$.

Since, the acceleration is upward for lift so, the accelerating force $ma$ acting on the block will be downwards.

From the free body diagram of the block, normal reaction on the body,

$N=m(g+a)\cos \theta$
$=1(10+2)\cos {30}^{\circ }$
$=6\sqrt{3}\text{N}$

Static friction:
$f=m(g+a)\sin \theta$
$=1(10+2)\sin {30}^{\circ }$
$=6\text{N}$

Weight of body:
$mg=1\times 10=10\text{N}$

Now, from equation of motion, distance travelled by the block,

$S=\frac{1}{2}a{t}^2=\frac{1}{2}\times 2\times 4^2=16\text{m}$

Now,
Work done by normal reaction force:
$W_N=NS\cos \theta =6\sqrt{3}\times 16\cos {30}^{\circ }=144\text{J}$

Work done by frictional force:
$W_f=f.S\sin \theta =6\times 16\times \sin {30}^{\circ }=48\text{J}$

Work done by gravity:
$|W_g|=mg.S=10\times 16=160\text{J}$

Work done by all the forces:
$W_{\text{net}}=W_g+W_N+W_f$
$=-160+144+48=32\text{J}$

Hence, option (c) is correct answer.