Question

# With the assumption of no slipping determine the mass m of the block which must be placed on the top of a 5 kg cart in order that the time period of oscillation is 2 sec. What is the minimum coefficient of static friction μs, for which the block will not slip relative to the cart, if the cart is displaced 300 mm, from the equilibrium position and released ?

A
m=2.08 kg, μs0.402
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B
m=1.079 kg, μs0.302
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C
m=2.08 kg, μs0.402
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D
m=1.079 kg, μs0.302
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Solution

## The correct option is D m=1.079 kg, μs≥0.302As we know that T=2π√5+m60 2=2π√5+m60 ⇒m=1.079 kg For maximum acceleration of SHM. amax=ω2A So maximum force an mass 'm' is m amax=mω2A. and this mω2A is provided by friction force only So μs mg ≥m ω2 A μs≥ω2Ag μs≥(2πT)2Ag ⇒μs≥(2π2)2×0.39.8 or μs≥0.302

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