A block of mass m = 3.5 kg is attached to a spring with spring constant k = 770

Question

A block of mass m = 3.5 kg is attached to a spring with spring constant k = 770N/m. It is initially at rest on an inclined plane that is at an angle of ? = 25

A block of mass m = 3.5 kg is attached to a spring with spring constant k = 770N/m. It is initially at rest on an inclined plane that is at an angle of ? = 25 A degree with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is ?k = 0.2. In the initial position, where the spring is compressed by a distance of d = 0.18 m, the mass is at its lowest position and the spring is compressed the maximum amount. Take the initial gravitational energy of the block as zero. a) What is the block's initial mechanical energy, E0 in J? b) If the spring pushes the block up the incline, what distance, L in meters, will the block travel before coming to rest? The spring remains attached to the block throughout its motion.

Explanation / Answer

The initial mechanical energy is purely due to the potential energy due to the spring,

E0 = 1/2 k d^2 = 12.5 J   [ANSWER]

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By work energy theorem,

Wncon = delta(PEg) + delta(KE) + delta(PEs)

where PEg = gravitational potential energy, and
PEs = potential energy due to the spring

Thus,

-uk M g L = -M g L sin 25 + 1/2 k (L^2 - d^2)

Plugging in our data and solving for L using the quadratic formula,

L = 0.190 m   [ANSWER]

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