A 7.85-kg block of ice, released from restat the top of a 1.75-m-long frictionle

Question

A 7.85-kg block of ice, released from restat the top of a 1.75-m-long frictionlessramp, slides downhill, reaching a speed 2.35 m/s of at the bottom.
(a) What is the angle between the ramp and thehorizontal?
1°

(b) What would be the speed of the ice at the bottom if the motionwere opposed by a constant friction force of 10.0 N parallel to thesurface of the ramp?
2 m/s (a) What is the angle between the ramp and thehorizontal?
1°

(b) What would be the speed of the ice at the bottom if the motionwere opposed by a constant friction force of 10.0 N parallel to thesurface of the ramp?
2 m/s

Explanation / Answer

A 8.40- block of ice, released from rest atthe top of a 1.07--long frictionless ramp, slides downhill,reaching a speed of 2.89 at the bottom.

What is the angle between the ramp and the horizontal?


What would be the speed of the ice at the bottom if the motion wereopposed by a constant friction force of 10.8 parallel to thesurface of the ramp? v^2 = u^2 + 2as
2.89^2 = 0^2 = 2a*1.07
a = 3.9

a = g sin(angle)
sin(angle) = 3.9/9.81
angle = 23.44 degrees

F = ma
F = 8.4 x 3.9 = 32.78 N (this is the force parallel to the ramp dueto gravity.

Friction = 10.8 N
Resultant force = 32.78 - 10.8 = 21.98 N
a = F/m = 21.98/8.4 = 2.617

v^2 = u^2 + 2as
v^2 = 0^2 + 2 x 2.617 x 1.07 = 5.601
v = 2.36657377 m/s
Three significant figures is appropriate:
v = 2.37 m/s v^2 = u^2 + 2as
2.89^2 = 0^2 = 2a*1.07
a = 3.9

a = g sin(angle)
sin(angle) = 3.9/9.81
angle = 23.44 degrees

F = ma
F = 8.4 x 3.9 = 32.78 N (this is the force parallel to the ramp dueto gravity.

Friction = 10.8 N
Resultant force = 32.78 - 10.8 = 21.98 N
a = F/m = 21.98/8.4 = 2.617

v^2 = u^2 + 2as
v^2 = 0^2 + 2 x 2.617 x 1.07 = 5.601
v = 2.36657377 m/s
Three significant figures is appropriate:
v = 2.37 m/s v^2 = u^2 + 2as
2.89^2 = 0^2 = 2a*1.07
a = 3.9

a = g sin(angle)
sin(angle) = 3.9/9.81
angle = 23.44 degrees

F = ma
F = 8.4 x 3.9 = 32.78 N (this is the force parallel to the ramp dueto gravity.

Friction = 10.8 N
Resultant force = 32.78 - 10.8 = 21.98 N
a = F/m = 21.98/8.4 = 2.617

v^2 = u^2 + 2as
v^2 = 0^2 + 2 x 2.617 x 1.07 = 5.601
v = 2.36657377 m/s
Three significant figures is appropriate:
v = 2.37 m/s

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