question about Work. please see attach.

Question

anonymous · Answer

attachment

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So what is the problem??

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please help to solve number 3b 4a and 4b.

anonymous · Answer

3b) The moment it loses contact the speed would be same when it was coming down and touching the spring. Since no friction  you can write euation for conservation of energy$$(1/2)mv^{2} = mgdsin(\theta)$$

anonymous · Answer

4a) calculate compression at equilibrium $$kx_{c} = mgsin{\theta}$$
the spring is further compressed from equilibrium by 21 cms so total compression is
$$X=x_{c} +21 cms$$
All this energy is spent against gravity and frictional force. So equation for conservation of energy is $$(1/2)kX^{2} = (mgsin{\theta} + mu*mgcos{\theta})l$$
Calculate $$l$$ from the above equation

anonymous · Answer

4b) The amount of energy lost as heat is work that is odne against friction$$Energy Lost as Heat  = \mu mglcos{\theta}$$
l u will get from 4a