A block of mass m=5kg is attached to a spring with a spring constant k=560N/m. I

Question

A block of mass m=5kg is attached to a spring with a spring constant k=560N/m. It is initially at rest on an inclined plane that is at an angle of ?=29 with respect to the horzontal, and coefficient of kinetic friction between the block and the plane is ?k=.16. In the initial position, where the spring is compressed by a distance of d=.12m, 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 i the block's initial mechanical energy, 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

a)

The initial energy possed by the block is only the elastic potential energy

(1/2)k x^2 = (1/2) (560)(0.12)^2 = 4.032 J

b)

The above calcualted energy moves the object along the inclined (work done in moving the object)

4.032 = ( m g sin 29 - u m g cos 29 ) L - (1/2) k L^2

4.032 = m g ( sin 29 - u cos 29 ) L - (1/2) k L^2

  4.032 = (5)(9.8) ( sin 29 - 0.16* cos 29 ) L  - (1/2)(560)L^2

Solving for L

L = 0.0935m

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