Q: A variable force F = 10 t is applied to block B placed on a smooth surface. The coefficient of friction between A & B is 0.5. (t is time in seconds. Initial velocities are zero, A is always on B)

(A) block A starts sliding on B at t = 5 seconds
(B) the heat produced due to friction in first 5 seconds is 312.5 J
(C) the heat produced due to friction in first 5 seconds is (625/8) J
(D) acceleration of A at 10 seconds is 5 m/s
Click to See Solution :
Ans: (A) , (D)
Sol:
fmax = μ × 3 g = 0.5 × 30 = 15 N
block A starts sliding when friction force becomes max. i.e. fmax = 15 at that instant (F.B. D.)

both will move with same acceleration
So , 15 = 3a
a = 5 m/s2
F – 15 = 7 a
10 t – 15 = 7 × 5
10 t = 50
t = 5 sec
Work done by friction in 5 seconds
$W = \int F.dS = \int 10 t .dS$
$W = \int_{0}^{5} 10 t .v dt $ ;Since $v = \int a dt = \int t dt $ & a = F/m = 10t/10 = t ;
$W = \int_{0}^{5} 10 t .\frac{t^2}{2} dt $
$W = \int_{0}^{5} 5 t^3 dt $
$W = 5[\frac{t^4}{4}]_{0}^{5}$
$W = \frac{625 \times 5}{4} J $