We want to load a 12kg crate into a truck by sliding it up aramp 2.5 m long, inc

Question

We want to load a 12kg crate into a truck by sliding it up aramp 2.5 m long, inclined at 30 degrees. A worker, giving nothought to friction, calculates that he can get te crate up theramp by giving it an initial speed of 5 m/s at the bottom andletting it go. But friction is not negligible; the crate slides 1.5m up he ramp, stops, and slides back down.

a. Assuming that the frictionforce acting onthe crate is constant, find its magnitude.

b. How fast is the crate moving when it reaches the bottom of the ramp? Show Work

Explanation / Answer

height of the incline = 2.5 sin 30 = 1.25 m

Let friction force be f

given initial speed = 5 m/sec

distance travelled = 1.5 m

final speed = 0

v^2 = u^2 + 2 * a * s

so, a = - 8.3333 m/sec2

gravitational force acting along incline = 12 * 9.8 * sin 30 = 58.8 N

net force down the incline = 58.8 + f

so net acceleration = (58.8 + f ) / 12 = 8.33333

so, f = 41.2 N

b)

when crate is moving down ,

net force down the incline = mg sin 30 - f = 58.8 - 41.2 = 17.6 N

so, acceleration = 17.6 / 12 = 1.4667 m/sec2

initial velocity = 0

acceleration = 1.46667 m/sec2

s = 1.5 m

v^2 = u^2 + 2 * a * s

so, v = 2.0976 m/sec

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