A 8 kg object slides down a ramp inclined at an angle of 12 degrees with respect

Question

A 8 kg object slides down a ramp inclined at an angle of 12 degrees with respect to the horizontal. The coefficient of kinetic friction between the object and the slope is 0.35. The object has an initial speed of 2.81303173534349 m/s as it slides down the slope. If the object slides down the slope a distance 1.1 m. determine the work done on the object by the force of gravity J, and the frictional force J. What is the object's initial kinetic energy? J. What is the object's final kinetic energy? J. What is the object's final speed? m/s. How far from the original starting point would the object slide before eventually coming to rest?

Explanation / Answer

here,

mass of object, m = 8 kg
Angle of inclened, A = 12 degrees
Coefficient of Kinetic Friction, uk = 0.35
Velocity, v = 2.813 m/s

distance sloped, d = 1.1 m

Part A:
Force due to gravity, Fg = mgCosA = 8 * 9.81 * COs12 = 76.765 J

Part B:
Frictional Force, Ff = uk*mg*CosA = uk*Fg = 0.35 * 76.765 = 26.868 J

Part C:
Initial KE = 0.5 * mv^2 = 0.5 * 8 * 2.813^2 = 31.652 J

Part D:
Final KE = Loss in Potential Energy = mg*h ( h is height of ramp)
KEf = mg*d*SinA
KEf = 8 * 9.81 * Sin12
KEf = 16.317 J

Part E:
Since KEf = 0.5 * m * vf^2

Solving for final speed, vf

vf = sqrt(KEf * 2 / m)

vf = sqrt(16.317 * 2 / 8)

vf = 2.02 m/s

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