A 10.4-kg block rests on a horizontal table and is attached to one end of a mass

Question

A 10.4-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 4.15 m/s in 1.91 s. In the process, the spring is stretched by 0.229 m. The block is then pulled at a constant speed of 4.15 m/s, during which time the spring is stretched by only 0.0855 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

a)______

b) ______

Explanation / Answer

1)

Acceleration = V/t = 4.15/1.91 = 2.17 m/s^2

Now, Force = mass * acceleration

or, F = 10.4*2.17 = 22.568 N

For spring F= K*X

OR, K = 22.568/0.229 = 98.55 N/m

2) After the block aquires constant speed, the spring force is equal to the frictional force.

Thus mu*m*g = K*X

or, mu*10.4*9.8 = 98.55 * 0.0855

or , mu = 0.0826

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