a block of mass 0,5kg is attached to a spring .The spring is compressed to a distance of0,1m.The block is released .The block moves 0,07m past the equilibrium position .What is the force of friction that does work on the block.

Question

spring constant is 500

Answer 1

We will first find the potential energy stored in the spring when it is compressed 0.1m, then we will calculate the work done by the force of friction as the block moves 0.07m past the equilibrium position.

1) Potential energy stored in the spring when compressed:

PE_spring = 0.5 * k * x^2
where k is the spring constant (500 N/m), and x is the compressed distance (0.1m)

PE_spring = 0.5 * 500 * (0.1)^2
PE_spring = 250 * 0.01
PE_spring = 2.5 J

2) Work done by the force of friction:

At the equilibrium position, all the potential energy stored in the spring is converted into kinetic energy. As the block moves past the equilibrium position, some of the kinetic energy is used to do work against the frictional force.

The block moves 0.07m past the equilibrium position so the remaining potential energy in the spring at this point is:

PE_spring_remaining = 0.5 * k * x^2
PE_spring_remaining = 0.5 * 500 * (0.07)^2
PE_spring_remaining = 250 * 0.0049
PE_spring_remaining ≈ 1.225 J

The work done by the frictional force is the difference between the initial potential energy and the remaining potential energy:

Work_friction = PE_spring - PE_spring_remaining
Work_friction = 2.5 - 1.225
Work_friction ≈ 1.275 J

Now we can find the force of friction:

Work = Force x Distance
Force_friction = Work_friction / Distance
Force_friction = 1.275 / 0.07
Force_friction ≈ 18.214 N

So the force of friction that does work on the block is approximately 18.214 N.