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Question

Consider the situation shown in figure (8-E2). The system is released from rest and the block of mass 1.0 kg is found to have a speed 0.3 m/s after it has descended through a distance of 1 m. Find the coefficient of kinetic friction between the block and the table.

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Solution

Given,

m1=4 kg, m2=1kg,

v2=0.3 m/sec

v1=2×(0.3)=0.6 m/sec

x1=2x2 m in this system

h = 1 m = height descent travelled by 1 kg block = x2

x1=2×1=2 m = distance travelled by 4 kg block

Taking u=0 and

Applying, change in K.E. = Work done (for the system)

(12)m1v21+(12)m2v22=(−μR)x1+m2gh

[R = 4g = 40 N]

⇒12×4×(0.36)+12×1×(0.09)=−μ×40×2+1×10×1

⇒0.72+0.045=−80μ+10

⇒μ=8.8380=0.11

73

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