Question

A block of ice mass 2.0 kg is pushed down an inclined plane of inclination ${37}^{\circ }$ with a force of 20 N acting parallel to the incline. It is found that the block moves on the incline with an acceleration of $10m/s^2$ If the block started from rest, find the work done$\left(W_f\right)$ by the applied force in the first second, $W_g$ by the weight of the block in the first second and $\left(W_f\right)$ by the frictional force acting on the block in the first second. Take $g=10m/s^2$

• A. W_F =100 J; W_g = 60 J; W_f = 60 J
• B. W_F =100 J; W_g = 60 J; W_f = -60 J
• C. W_F =100 J; W_g = 100 J; W_f = 0 J
• D. None of these

Answer 1

The correct option is B

W_F =100 J; W_g = 60 J; W_f = -60 J

So question is asking for work done in the first second

$a=10\frac{m}{s^2}$

$u=0\frac{m}{s}$

$t=1sec$

So, in 1 sec displacement of block will be $s=ut+\frac{1}{2}a{t}^2$

s = 5m

$W_F$ = Work done by force = $F\times s\times cos\theta$

= $20\times 5\times cos{0}^{\circ }$

= 100 J

$W_g$ = Work done by weight = $mg\times s\times cos\theta$

=$2\times g\times 5\times cos{53}^{\circ }$

= 60 J

$W_f$ = Work done by friction = $f_r\times s\times cos\theta$

Oh, so we need NLM to find friction from free body drawing of object.

$F+mgcos{53}^{\circ }-f_r$ = $ma$

$20+12-f_r=2\times 10$

$f_r=12N$

=$W_f=12\times 5\times cos{180}^{\circ }=-60J$