A block of mass m = 3.5 kg is attached to a spring with spring constant k = 890

Question

A block of mass m = 3.5 kg is attached to a spring with spring constant k = 890 N/m. It is initially at rest on an inclined plane that is at an angle of theta = 26 degrees with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is uk = 0.15. In the initial position, where the spring is compressed by a distance of d = 0.11 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. (ANSWERED) What is the blocks initial mechanical energy, E0 in J? Answer: 5.385 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 both the block and the fixed wall throughout its motion.

Explanation / Answer

According to the given problem,

a) Intial mechanical energy,

½kx² = ½*890*0.11² = 5.3845J

b)Using the energy conservation principal,

The weight (force) in line with the incline = 3.5*9.8*sin26 = 15.036N
The force of friction in line with the incline (opposing motion) = 0.15*3.5*9.8*cos26 = 4.6243N

So 5.3845J = [15.036 + 4.6243]*L

L = 0.274 m = 27.4 cm

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