An object with a mass of 12 kg is initially at rest at the top of a frictionless

Question

An object with a mass of 12 kg is initially at rest at the top of a frictionless inclined plane that rises at 30 degrees above the horizontal. At the top, the object is initially 8.0 meters from the bottom of the inclined, as shown in the figure. When the object is released from this positon, it eventually stops at a distance d from the bottom of the inclined plane along a horizontal surface, as shown. The coefficicent of kinetic friction between the horizontal surface and the object is 0.10, and air resistance is negligible. Find the distance d.

Explanation / Answer

here,

mass of the object , m = 12 kg

theta = 30 degree

l = 8 m

let the velocity at the bottom be u

using conservation of energy

0.5 * m * u^2 = m * g * ( l * sin(theta))

0.5 * u^2 = 9.8 * 8 * sin(30)

u = 8.85 m/s

the accelration due to friction , a = uk * g

a = - 0.98 m/s^2

using third equation of motion

v^2 - u^2 = 2*a*d

0 - 8.85^2 = - 2*0.98 * d

d = 39.96 m

the distance d is 39.96 m

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