Question

A 1.1 kg block with a speed of 2.1 m/s travels down a frictionless table and hits a horizontal spring attached to a wall. The spring constant is 980 N/m.

a. What is the maximum compression of the spring?

b. What is the maximum elastic force exerted by the spring on the block?

Please show all steps in the calculation.

Answer 1

The block of mass 1.1 Kg is moving with speed of 2.1 m/s , move down and hits a spring.

In this process the kinetic energy of the block is used in compressing the spring.

Kinetic energy of the block is given as,

.. .. ... ... (1)

Potential Energy of the spring is given as,

.. .... . .. .... (2)

Using principle of conservation of energy.,

K. E of block = P. E of spring stored due to moving mass

  

We have m= 1.1 kg , v = 2.1 m/s , k = 980 N/m

So maximum compression in spring is given as,

= 2.1 × 0.03350

meter

... .. This is the value of the maximum compression in spring. .

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Part b)

Maximum elastic force exerted by spring on block is,

= 980 × 0.07035 N

= 68.94 Newton

So max force exerted by the spring on block is 68.94 N.

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