Near the surface of the Earth, an ideal spring with spring constant k = 50 N/m i

Question

Near the surface of the Earth, an ideal spring with spring constant k = 50 N/m is on a frictionless horizontal surface at the base of a frictionless inclined plane as shown in the figure. A block with mass M = 0.5 kg is pressed against the spring, compressing it 0.5 m from its equilibrium position. The block is then released and is not attached to the spring. If the block slides a distance d = 1.5 m up the inclined plane before coming to rest and then sliding back down, what is the angle ? of the inclined plane relative to the horizontal?

The answer is 58.2

Explanation / Answer

It is simply conservation of energy, let the angle be ? , so height of fall = 1.5sin? m

now mgh=0.5mv^2=0.5kx^2

0.5*9.8*1.5sin? = 0.5* 50*0.25

?=58.116 degrees

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