Question

A 8.00-kg block of ice, released from rest at the top of a 1.08-mm-long frictionless ramp, slides downhill, reaching a speed of 2.70 m/s at the bottom.

A: What is the angle between the ramp and the horizontal?

B: What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.1 N parallel to the surface of the ramp?

Answer 1

given
m = 8 kg
d = 1.08 m
vi = 0
vf = 2.7 m/s
A) let theta is the angle of inclination and a is the acceleration of the block.

use, vf^2 - vi^2 = 2*a*d

a = (vf^2 - vi^2)/(2*d)

= (2.7^2 - 0^2)/(2*1.08)

= 3.38 m/s^2

now use, a = g*sin(theta)

sin(theta) = a/g

theta = sin^-1(a/g)

= sin^-1(3.38/9.8)

= 20.2 degrees <<<<<<<-------------Answer

B) let a is the acceleration of the block,

Net force acting on the block along the ramp,

Fnet = (m*g*sin(theta) - friction)

m*a = (m*g*sin(theta) - friction)

a = (m*g*sin(theta) - friction)/m

= (8*9.8*sin(20.2) - 10.1)/8

= 2.12 m/s^2

now use, vf^2 - vi^2 = 2*a*d

vf^2 - 0^2 = 2*2.12*1.08

vf = sqrt(2*2.12*1.08)

= 2.14 m/s degrees <<<<<<<-------------Answer