Problem 8.42 A spring (70 N/m ) has an equilibrium length of 1.00 m . The spring

Question

Problem 8.42

A spring (70 N/m ) has an equilibrium length of 1.00 m . The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 with respect to the horizontal. The spring is t

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

hen released. (Figure 1)

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 ?

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.

m

Problem 8.42

A spring (70 N/m ) has an equilibrium length of 1.00 m . The spring is compressed to a length of 0.50 m and a mass of 2.2 kg is placed at its free end on a frictionless slope which makes an angle of 41 with respect to the horizontal. The spring is t

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

hen released. (Figure 1)

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 ?

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.

m

Explanation / Answer

part B :

apply the law of conservation of energy as EPE = PE

i,e 0.5 kx^2 = m g h

h = L2 sin 41

0.5kx^2=mg*L sin 41

L = (0.5 * 70 * 0.5* 0.5)/(2.2 * 9.8 *sin 41)

L = 0.618 m

-------------------------------------

part C: use the formula for coefficient of friction

u mg cos theta x = 0.5 kx^2

u = (0.5 * 70 * 0.5* 0.5)/(2.2 * 9.8 * cos 41 * 0.5)

u = 1.075

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