Question

A block of mass m = 3.5 kg is attached to a spring with spring constant k = 520 N/m. It is initially at rest on an inclined plane that is at an angle of θ = 21° with respect to the horizontal, and the coefficient of kinetic friction between the block and the plane is μ_k = 0.16. In the initial position, where the spring is compressed by a distance of d = 0.14 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.

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.

Answer 1

MASS =m = 3.5 Kg ; spring constant = 520N/m ; initial velocity of mass = 0m/s ; Angle of inclined plane = 21deg; =0.16 , Spring compression at initial position = 0.14 m ( at maximum compression -with m at lowest position where Gravitational PE is taken as 0 Joules)

The initial PE in spring = 0.5 K x^2 = 0.5 x 520 N/m x (0.14m)^2 = 5.10 Joules

Initial KE = 0 since mass is at rest, Initial Gravitational PE = 0 Joules

Total mechanical energy in Initial position = 5.10 J + 0 J + 0 J = 5.10 J

Let Block move to a final position such that the mass would have traveled distance L along the inclined plane and comes to rest there.

The mass m would have increased it's height from initial position by distance

At this position the spring initial PE would have completely converted into gravitational PE of mass m and heat energy due to friction.

The heat energy lost by way friction =    0.16 x 3.5 Kg x 9.8 m/s^2 x cos 21 x L = 5.123 L

The PE gain by mass . = 3.5 Kg x 9.8 m/s^2 x 0.358 L = 12.29 L joules

KE at final position = 0 J

Final Mechanical Energy = 12.3 L Joules

Initial mechanical energy - frictional energy loss = Final Mechanical Energy

5.10 J - 5.12 L Joules = 12.3 L Joules . or . 5.10 J = 17.42 L or L = 5.10 /17.42 = 0.29 meters

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L sin = Lsin21 = 0.358L

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