A block is released from rest on a frictionless incline as shown. It slides down

Question

A block is released from rest on a frictionless incline as shown. It slides down 4.00 m before getting in contact with a spring. The mass of the spring is negligible. How much distance the spring is compressed before the block comes to a stop? After the block comes to a stop, the spring pushes it back up and eventually looses contact with the block. At the moment when the block looses contact with the spring. What is the speed of the block? As the block continues moving upward, what is the maximum distance the block travels up, measured from the position when it first looses contact with the spring?

Explanation / Answer

conservation of energy is used.

mg(h+x) sin53.1  = 1/2 Kx^2 

total energy lost by the body = total energy gained by the spring

solving, we get,

32(4+x) = K x^2 

the value of k is required to do this

speed of block = 

energy of spring = kinetic energy of the block 

1/2 Kx^2 = 1/2 mv^2  = 

63.97 = 1/2 mv^2 

v= 7.99 m/s

since there is no loss in the system , the block reaches 4 mts above the point of contact with the spring

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