Write your answers clearly. Show all steps. This is an open book quiz. Do not co

Question

Write your answers clearly. Show all steps. This is an open book quiz. Do not copy the answers from textbooks, online solution banks, or from other students. Your grade will be based on how you arrived at the solution with all intermediate steps. Due on Monday, Jun 22, 9:00 AM. Late submissions will not be accepted. 1. A 2.0 kg block Is released from rest at the top of 22 frictionless inclined plane of height 0.65 m. At the bottom of the inclined plane, it collides with and sticks to a block of mass 3.5 kg. The two blocks slide a distance of 0.57 m across a horizontal plane before coming to a rest. What is the coefficient of friction of the horizontal surface?

Explanation / Answer

here,

m1 = 2.0 kg

m2 = 3.5 kg

theta = 22 degree

h = 0.65 m

let v1 is the velocity of block 1 at the bottom on incline

using conservation of energy

change in potential energy = change in kinetic energy

m1 * g * h = 0.5 * m1 * v1 ^2

2 * 9.8 * 0.65 = 0.5 * 2 * v1^2

v1 = 3.56 m/s

velocity of sticked blocks be Vf

using conservation of momentum

Vf * (m1 + m2) = m1 * v1

Vf * (2+3.5) = 2 * 3.56

Vf = 1.29 m/s

let u is the coefficient of friction

and using work enegy theorm

work done by friction force = change in kinetic energy

u * m * g * d = 0.5 * m * Vf^2

u * 9.8 * 0.57 = 0.5 * 1.29^2

u = 0.148

the cofficient of friction of the horizontal surface is 0.148

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