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Q.

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/s2)

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a

Block will break off the plank before relative motion starts.

b

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

c

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

d

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

answer is B.

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Detailed Solution

since R= Neff=mg- F Sin370 =0.5×10-5 t ×35R=5-3t For breaking off (R=0)t=5/3secfL=μR=110[5−3t]=12−310t For 0.5kg block, 4t−f=0.5aFor 1kg block,f=1aSolving both, we getf=8t3Relative motion starts when f>fL⇒8t3>12−3t10⇒t>1589s

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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/s2)