A block of mass 4kg is placed in contact with the front vertical surface of a lorry. The coefficient of friction between the vertical surface and block is 0.8. The lorry is moving with an acceleration of 15 m/s2 (g = 10ms-2). The force of friction between lorry and block is
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a
48N
b
24N
c
40N
d
zero
answer is C.
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Detailed Solution
fs = µsma = 0.8 x 4 x 15 = 48Nmg = 4 x 10 = 40N as mg < fsfriction = mg = 40 N