A spring (80N/m ) has an equilibrium length of 1.00 m. The spring is compressed

Question

A spring (80N/m ) has an equilibrium length of 1.00 m. The spring is compressed to a length of 0.50 m and a mass of 1.9kg is placed at its free end on a frictionless slope which makes an angle of 41 ? with respect to the horizontal. The spring is then released. (Figure 1)

Part A.

If the mass is not attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?

Express your answer using two significant figures.

Part B.

If the mass is attached to the spring, how far up the slope from the compressed point will the mass move before coming to rest?

Express your answer using two significant figures.

Part C.

Now the incline has a coefficient of kinetic friction ?k. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction ?k?

Express your answer using two significant figures.

Explanation / Answer

1.

In this first problem the system starts with elastic potential energy: EPE=1/2*k*dx^2.

As the mass slides up the incline the elastic energy is converted into gravitational potential energy:

GPE=m*g*dh

where dh is equal to the distance L up the incline that the mass slides multiplied by the sine of the angle Q of the incline: dh=L*sin(Q)

Using energy conservation make these two quantities equal:

EPE=GPE which becomes 1/2*k*dx^2=m*g*L*sin(Q)
2

. If the mass is attached to the spring at the highest point there will be BOTH elastic and gravitational energy and the energy conservation equation becomes:

EPEo=GPE+EPEf which becomes 1/2*k*dx^2=m*g*L*sin(Q)+1/2*k*[L-dx]^2
3

. Finally, in this third case the original elastic energy converts into both gravitational energy plus work done against the force of friction:
EPE=GPE+W which becomes 1/2*k*dx^2=m*g*dx*sin(Q)+Ff*dx

where the force of friction will be equal to the product of the normal force Fn=m*g*cos(Q) acting on the mass and the

corresponding coefficient of friction mu:

1/2*k*dx^2=m*g*dx*sin(Q)+Ff*dx which becomes 1/2*k*dx^2=m*g*dx*sin(Q)+m*g*cos(Q)*mu*dx

that all the distances in this version of the problem are the same dx since the mass slides until it reaches the equilibrium position of the spring as given

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