Question

In: Physics

A 16.3-kg block rests on a horizontal table and is attached to one end of a...

A 16.3-kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of 5.99 m/s in 1.37 s. In the process, the spring is stretched by 0.180 m. The block is then pulled at a constant speed of 5.99 m/s, during which time the spring is stretched by only 0.0584 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

Can someone use these numbers, step by step with equations?

Expert Solution

A block of mass m = 16.3 kg

The block is accelerating from rest to a speed 5.99 m/s in time interval 1.37s.

Thus, the acceleration of the block is

(a) Apply Newton laws, when block is accelerating:

...... (1)

Here, m is the mass of the block, a is the acceleration of the block, is the coefficient of friction, k is the spring constant, and x₁ is the compression of the spring in frist case.

Apply Newton laws, when the block is moving with constant velocity:

(since it is moving with constant velocity)

Here, x₂ is the compression of the spring in second case.

Thus, the coefficient of kinetic friction is

Substitute the coefficient of friction value in equation (1).

Therefore, the spring constant is,

(b) The coefficient of kinetic friction is given by

genius_generous answered 2 years ago

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