Question

A small block of mass 20 kg rests on a bigger block of mass 30kg , which lies on smooth horizontal plane. Initially the whole system is at rest. The coefficient of friction between the blocks is 0.5 . A horizontal force F=50N is applied on the lower block. Find the work done by frictional force on upper block in t=2sec

Answer 1

Final Answer : 40N (with respect to ground)

Steps:
1) Assume both the blocks move together, i.e,there is no relative motion.
Then,
Take blocks as system, :
$F_{ext} = 50N \\ => (20+30) a = 50 \\ => a = 1 m/s^2$

Therefore, System will move in 2 s :
$S =ut + \frac{1}{2}at^2 \\ => 0+ \frac{1}{2}*1*2^2 = 2 m$

2) Static Friction will act on upper block which helps in moving forward.
FBD of 20 kg block is shown in pic 2 .
We can say,
$f = ma \\ => f = 20*1 = 20N$

$f_{max} = \mu N = \mu *mg \\ => f_{max} = 0.5 * 20*10 = 100N$

3) We observed,
$f_{max} > f$ .
=> Our Assumption is right.

4) Work depends on Frame of Reference.
Displacement = 2 m
Static Frictional Force = 20N
Work Done by frictional force on 20kg block wrt ground = 20* 2 = 40J

Here, Interesting Part :
Work Done by Friction is positive.