Question

A 25.6 kg. block sits on a 55 degree ramp with a coefficient of static friction of 0.88 and a coefficient of kinetic friction of 0.41. Inititally the block is 1.7m above the ground. The ground and the ramp are made of the same material. Find out how far from the end of the ramp the block slides.

Answer 1

here,

theta = 55 degree

mass , m = 25.6 kg

us = 0.88 , uk = 0.41

let the final speed be v

work done by friction = initial potential energy - final kinetic energy

uk * m * g * cos(theta) = m * g * h - 0.5 * m * v^2

0.41 * 9.81 * cos(55) = 9.81 * 1.7 - 0.5 * v^2

solving for v

v = 5.36 m/s

accelration due to friction , a = - uk * g

a = - 0.41 * 9.81 = - 4 m/s^2

let the distance travelled be s

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

0 - 5.36^2 = - 2 * 4 * s

s = 3.57 m

the block slides 3.57 m after the end of the ramp