Question

In the situation shown in figure, coefficient of friction between the block and ground is $0.2,m=2kg,F_1=5N$ and $F_2=4N$. If the block is displaced by $10m$, match the following. $(g=10m{s}^{-2})$
Column I Column II
(A) Magnitude of work done by $F_1$ (p) $20J$
(B) Magnitude of work done by $F_2$ (q) $23.07J$
(C) Magnitude of work done by friction (r) $46.93J$
(D) Magnitude of increase in K.E. of the block (s) $50J$

• A. A $\to$ (r); B $\to$ (q); C $\to$ (p); D $\to$ (s)
• B. A $\to$ (p); B $\to$ (r); C $\to$ (s); D $\to$ (q)
• C. A $\to$ (s); B $\to$ (p); C $\to$ (r); D $\to$ (q)
• D. A $\to$ (q); B $\to$ (r); C $\to$ (p); D $\to$ (s)

Answer 1

Answer: (C)

A $\to$ (s); B $\to$ (p); C $\to$ (r); D $\to$ (q)

Work done by $F_1,W_1=F_{1}S=5\times 10=50J$

Work done by $F_2,W_2=F_{2}S\cos 60=4\times 10\times \frac{1}{2}=20J$

Work done by friction, $W_3=-\mu Ns$

$=-\mu (mg+F_{2}sin60)s$

$=-0.2\times (2\times 10+4\times \frac{\sqrt{3}}{2})\times 10$

$=-(40+4\sqrt{3})J$
$\therefore$ Change in K.E. = Net work done
$=W_1+W_2-W_3=30-4\sqrt{3}J$