Question

Constants PartA A crate of sports equipment, with a mass of 8.40- kg, is released from rest at the top of a 1.46-m- long ramp. The ramp has small wheels on it so that the crate slides down the ramp without any friction, reaching a speed of 2.79 m/s at the bottom. What is the angle between the ramp and the horizontal? Submit Request Answer Part B What would be the speed of the crate at the bottom of the ramp if the motion were opposed by a constant friction force of 10.5 N parallel to the surface of the ramp? m/s Submit Request Answer

Answer 1

Cc atc Ramp

Gravitational acceleration = g = 9.81 m/s²

Length of ramp = L = 1.46 m

Mass of the crate = m = 8,4 kg

Speed of the crate at the bottom of the ramp = V = 2.79 m/s

Initial height of the crate = H

Angle the ramp makes with the horizontal = $\theta$

From the figure,

H = LSin$\theta$

No friction is present.

By energy balance the potential energy of the crate at the top is converted to kinetic energy at the bottom.

mgH = mV²/2

gH = V²/2

gLSin$\theta$ = V²/2

(9.81)(1.46)Sin$\theta$ = (2.79)²/2

Sin$\theta$ = 0.2717

$\theta$ = 15.76^o

H = LSin$\theta$

H = (1.46)Sin(15.76)

H = 0.397 m

Now a constant friction force of 10.5 N acts on the crate parallel to the surface of the ramp.

Friction force = f = 10.5 N

Speed of the crate at the bottom of the ramp = V₁

By energy balance the initial potential energy of the crate is converted to friction energy and kinetic energy of the crate.

mgH = fL + mV₁²/2

(8.4)(9.81)(0.397) = (10.5)(1.46) + (8.4)(V₁²)/2

V₁² = 4.139

V₁ = 2.03 m/s

A) Angle between the ramp and the horizontal = 15.76^o

B) Speed of the crate at the bottom of the ramp when friction is acting on the crate = 2.03 m/s