A 4 kg block moves on a horizontal, frictionless surface and collide with a spri
ID: 1492354 • Letter: A
Question
A 4 kg block moves on a horizontal, frictionless surface and collide with a spring of spring constant k that is fixed to a wall. When the bloc momentarily stops, the spring has been compressed by 0.20 m. After rebounding, the block has a speed of 1.00 m/s. Next, the spring is put on a inclined surface with its lower end fixed in place (Fig-8-30). The same block i now released on the incline at a distance of 5.0 m from the spring's free enc When the block momentarily stops, the spring has been compressed by 0.3 What is the coefficient of kinetic friction between the block and the incline How far does the block then move up the incline from the stopping point?Explanation / Answer
To find spring constant
=> 1/2 * k * 0.20 * 0.20 = 1/2 * 4 * 1 * 1
=> k = 100 N/m
a) Now, on the incline
=> 4 * 9.8 * sin30 * 5 - mu * 4 * 9.8 * cos30 * 5 = total energy
=> 4 * 9.8 * sin30 * 5 - mu * 4 * 9.8 * cos30 * 5 = 1/2 * 100 * 0.3 * 0.3 + mu * 4 * 9.8 * cos30 * 0.3
=> 93.5 = 179.925 * mu
=> coefficient of kinetic friction = 0.5196
b) Here, 1/2 * 100 * 0.3 * 0.3 = 0.5196 * 4 * 9.8 * cos30 * x + 4 * 9.8 * sin30 * x
=> 4.5 = 37.24x
=> x = 0.1208 m
=> distance block moves up the incline . = 0.1208 + 0.3 = 0.4208 m