Question

Question 7 7 of 7 Constants Part A A 9.00-kg block of ice, released from rest at the top of a 1.57-m-long frictionless ramp, slides downhill, reaching a speed of 2.93 m/s at the bottom What is the angle between the ramp and the horizontal? AX Submit Request Answer Part B What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 11.0 N parallel to the surface of the ramp? A2 m/s Submit Request Answer Return to Assignment Provide Feedback

Answer 1

part A)

let the angle of the ramp is theta

Using conservation of energy

0.50 * m * v^2 = m * g * L * sin(phi)

0.50 * 2.93^2 = 9.8 * 1.57 * sin(phi)

solving for phi

phi = 16.2 degree

the angle between the ramp and horizontal is 16.2 dgeree

part B)

let the velocity is v

Using work energy theorem

0.50 * 9 * v^2 = 9 * 9.8 * 1.57 * sin(16.2 degree) - 11 * 1.57

solving for v

v = 2.18 m/s

the speed at the bottom is 2.18 m/s