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Asked by Prashant DIGHE | 17 Sep, 2019, 10:02: PM
Expert Answer
Figure shows the different forces acting on block of mass m placed on inclined plane.
when block of mass remains stationary, if block of mass M is released so that it comes down by x,
then by conservation of energy, we have ΔK + ΔU = 0 + (ΔU)spring + (ΔU)M
Hence (1/2)kx2 - ( M g x ) = 0 or spring force k x = 2 M g
To move the block of mass m upside of inclined plane, let us apply Newton's law to block of mass m .
we have, ( k x ) ≥ mg sin37 + μ mg cos37 .................(1)
2 Mg ≥ mg (3/5) + (3/4) mg (4/5)
M ≥ (3/5)m
Answered by Thiyagarajan K | 18 Sep, 2019, 07:38: PM
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