A block of mass 0.5 kg is placed over a long plank m of mass 1 kg. The friction coefficient between block and plank is 0.1. The system is placed over a smooth horizontal surface. A time varying force F = 5t Newton starts acting on the block as shown in figure. ${(g =10 m/s}^{2})$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Block will break off the plank before relative motion starts.

Block will break off the plank after the start of relative motion.

Relative motion between block and plank starts at t = 15/89 sec.

Relative motion between block and plank starts at t = 11/89 sec.

answer is B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} \sin c e R = N_{e f f} = m g - F S i n 37^{0} = 0.5 \times 10 - 5 t \times \frac{3}{5} \\ R = 5 - 3 t \\ For breaking off (R=0) \\ t = 5 / 3 \sec \\ f_{L} = μ R = \frac{1}{10} [5 - 3 t] = \frac{1}{2} - \frac{3}{10} t \end{array}$

$\begin{array}{l} F o r 0.5 k g b l o c k, \\ 4 t - f = (0.5) a \\ F o r 1 k g b l o c k, \\ f = (1) a \\ S o l v i n g b o t h, w e g e t \\ f = \frac{8 t}{3} \\ R e l a t i v e m o t i o n s t a r t s w h e n \\ f > f_{L} \\ \Rightarrow \frac{8 t}{3} > \frac{1}{2} - \frac{3 t}{10} \\ \Rightarrow t > \frac{15}{89} s \end{array}$