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Asked by Prashant DIGHE | 17 Sep, 2019, 22:02: PM
answered-by-expert 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, 19:38: PM

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