A 13.2-kg block rests on a horizontal table and is attached to one end of a mass

Question

A 13.2-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 4.17 m/s in 1.01 s. In the process, the spring is stretched by 0.162 m. The block is then pulled at a constant speed of 4.17 m/s, during which time the spring is stretched by only 0.0589 m. Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

Explanation / Answer

This is definitely not simple harmonic motion because the motion is not periodic, the block does not move back and forth about some fixed point, its acceleration is constant and there is friction. In short, it does not obey the equation "a = -w^2 x".

To solve this problem, you must apply Newton's 2nd Law.

Net force, F = ma

(a) For a spring-mass system, force needed to extend a spring is given by F = kx, where k is the spring constant and x is the extension in the spring.

The block is initially at rest. Acceleration is uniform,

so a=(4.17-0)/1.01= 4.123 m/s^2.

ma = kx

Spring constant, k = ma/x = 13.2 * 4.123 / 0.162 = 336 N/m.

(b) In the 2nd case, the block moves with constant speed, i.e. a=0, so the net force F=0.

Let the force to stretch the spring = F_1, the frictional force = f.

F = F_1 - f = 0 => f = F_1

f = F_1 = kx = 336 * 0.0589 = 19.7904 = 19.8 N

f = uR, where u is the coeff of kinetic friction, R is the normal reaction force the table exerts on the block. R is perpendicular to the table surface.

R = mg = 13.2 * 9.8 = 129.36 N.

Coeff, of kinetic friction, u = f/R = f/mg = 19.8/129.36 = 0.15306 = 0.153

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