Question

A block of mass $m=1\text{kg}$ is at rest with respect to a rough wedge as shown.The wedge starts moving up from rest with an acceleration of $a=2{m/s}^2$ and the block remains at rest with respect to wedge. Then in 4 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 CORRECT 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 $A\to R,B\to P,C\to S,D\to Q$

$A\to R,B\to P,C\to S,D\to Q$
$N=m(g+a)\cos \theta =6\sqrt{3}\text{N}$
Static friction
$f=m(g+a)\sin \theta =6\text{N}$
$mg=10\text{N}$
Now, $S=\frac{1}{2}a{t}^2=16\text{m}$
$\therefore W_N=N.S\cos \theta =144\text{J}$
$W_f=f.s\sin \theta =48\text{J}$
$|W_g|=mg.s=160\text{J}$
$W_{\text{net}}=W_g+W_N+W_f$
$=-160+144+48=32\text{J}$