Need some help in this. Would appreciate steps.

Question

A spring attached to 2 kg block is compressed by 20 cm and then released. If the

coefficients of friction between the block and the floor are μs = 0.52 and μk = 0.38,

determine how far does the block slide on the floor.

Answer 1

Here is a solution to the problem symbolically, meaning I will not substitute in the values for any of the variables, but I will solve the problem for any given inputs

m = mass of the block

d = compression of the spring

k = spring constant

us = coefficient of static friction

uk = coefficient of kinetic friction

N = normal force

Ff = Frictional force

g = acceleration due to gravity

x = how far the block slides from its initial point, compressed against the spring.

We must first determine if the block actually slides when the spring is compressed, because if the force of the spring is not strong enough to overcome the force of static friction, the block will not move at all, and x=0. First do a free body diagram on the block

The diagram will show the weight mg down, the normal force N up, the spring force kd to the right, and the frictional force Ff to the left.

To check to see if the block remains still, check to see if the force exerted by the string exceeds the maximum possible static friction force

kx ?> Ff = us*N

Because the block is not accelerating vertically, the vertical forces must cancel, meaning

N-mg = 0 => N=mg

this means us*N = us*m*g

So the block will move if k*d>us*m*g

First, check to see if this inequality holds. If it does not, then static friction keeps the block in place and x=0

If it does, then the block slides. How far it slides can be figured out by conservation of energy.

There is still no acceleration in the vertical direction, meaning that N=mg. However, the friction force is now kinetic, so Ff = uk*m*g.

The net work done over the trajectory is equal to the change in kinetic energy of the block. Since the block starts and ends at rest, its initial and final kinetic energies are 0. Therefore, the net work is 0.

∑W = 0

We must now consider the forces that do work. The block slides horizontally, meaning that the vertical gravitational and normal forces do no work. The only forces that do work are the spring force and the friction force.

Ws + Wf = 0 (Ws is the work of the spring force, and Wf is the work of the friction force)

The work of the spring is the change in potential energy of the spring. The spring starts out with a potential energy of (1/2)*k*d^2 and ends with a potential energy of 0, since it fully decompresses. This work is positive because it acts in the direction of motion.

(1/2)*k*d^2 + Wf = 0

Since the friction force is constant, its work is the force times the distance over which it acts. Since friction acts for the entire time the block is moving, the distance is the travel distance we are trying to find, x. The friction work is negative, because it acts against the direction of motion

Wf = -Ff*x = -uk*m*g*x

Substituting gives

(1/2)*k*d^2 - uk*m*g*x = 0

Since the only unknown is x, it can be solved for

uk*m*g*x = (1/2)*k*d^2

x=(k*d^2)/(2*uk*m*g)

In summary, there are two cases to consider

1) if k*d<us*m*g

x=0

2) k*d>=us*m*g

x=(k*d^2)/(2*uk*m*g)

From here, you can get the answer by substituting in the given values. However, you do need to know what the spring constant is to find the answer in each case. If the problem did not give you a spring constant, then it did not give you enough information to solve the problem.