Question

A crate is moving on a conveyor belt at a speed of 5 ft/s when it is delivered at point A to a ramp with an angle of 30 degrees. If the kinetic coefficient between the crate and the ramp is 0.20, find the speed of the crate when it reaches point B. Report your answer in ft/s. 5 ft/s A 20 Ft B

Answer 1

To solve this question, You should know how to draw free body diagram and basic velocity and acceleration equation or concepts. In below figure, I have drawn starting position and when its reached some distance from point A and FBD of crate.

Vi = 5 ft/s * FBD of crate N = Normal fk Reaction Kinetic 20 ft for faction force U 0-30 B 30 ngensch mesine 30DB

From free body diagram, we get two equations as written below by equating the forces acting in opposite direction. (Assuming you know concept and understood from the figure)

1) $m*g *cos\theta = N$ (N is normal reaction acting on crate by ramp)

2)

m * g * sino - fk = = m * a

a is acceleration of crate

$f_{k}$ is frictional force which is equal to
2
where $\nu$ is kinetic fricion coefficient (crate is in motion so kinetic)

$\theta$ = 30. (See fig I have drawn)

Now, Put normal reaction value in second equation.

$m*g *sin\theta - \nu*N = m*a$

\$m*g *sin\theta - \nu* m*g *cos\theta = m*a$

we can remove m from both the side. Now put values and calculate a. (g = 32.174 ft/s²)   

(32.174 * sin 30) - (0.2 * 32.174 * cos 30) = a

a = 10.514 ft/s²

Now, from constant acceleration equations,

$S = V_{i}*t + (\frac{1}{2})* a*t^{2}$

Find time t from above equation by putting S = 20 ft. This is the time to reach at point B from point A.

20 = (5 * t) + [(10.514 * t^2)/2]

5.257 t^2 + 5 t - 20 = 0

Solve this equation and you will get

t = 1.53 seconds

Now, put this t value in

Vf= 5 + (10.514 * 1.53)

Velocity of crate at point B Vf = 21.086 ft/s.