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

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Ans: (A) , (D)

Sol:

f_{max} = μ × 3 g = 0.5 × 30 = 15 N

block A starts sliding when friction force becomes max. i.e. f_{max} = 15 at that instant (F.B. D.)

both will move with same acceleration

So , 15 = 3a

a = 5 m/s^{2}

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 $