- #1

Zanychap

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## Homework Statement

The situation given in the following question doesn't look possible to me.

A 1kg block situated on a rough incline is connected to a spring of spring constant 100 N/m. The block is released from rest with the spring in the unstretched position. The block moves 10cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has negligible mass and the pulley is frictionless. The angle of inclination is 37degree.

The answer(as given in the book) is 0.125, which can be obtained by the following:

**kx**(work-energy theorem)

^{2}/2=mgx(sin37-f*cos37)where k=100,x=0.1m,m=1kg,f=coefficient of kinetic friction

## Homework Equations

So, mgsin37=1*9.8*0.6=5.88N, and remains constant throughout the descent.

The spring force increases throughout the distance. At the end of the descent, the force exerted by the spring is kx=100*0.1=10N.

10>5.88. In addition, it is that 10N is aided by friction.

So, how can the spring force be greater than mgsin37 and still the body move downward upto that 10cm?

And the equation of forces when the block is at rest is:

kx+fmg*cos37=mgsin37

where f is obviously the coefficient of static friction

After putting the values, you will get f=-0.52, which is impossible.

## The Attempt at a Solution

Because of these facts, the situation described in the question doesn't seem probable.

Is my assessment correct?