a worker on a flatbed railroad car is pusging a crate. the train is moving at a

Question

a worker on a flatbed railroad car is pusging a crate. the train is moving at a constant speed of 15.0 m/s . the crate has a mass of 12kg, and in being pushed forward ovver a distance of 2.4m its velocity is increased (relative to the car) at constant accleration from rest to 1.5 m/s. show that the work done-kinetic energy theorem and implise-linear momentum theroem are valid for each observer (inside the train and on the ground.)

I know that we are goning to use the Formula W = KE = Fd = m(vf)2/2 - m(vi)2/2.. and the other one will be J=P = Ft=mvf=mvi, but i'm not sure if we should use the same accelration for both the train and the ground weather it woulf be different, and how to choose the velocity given.

Explanation / Answer

Answer:

Here,

Train Velocity (vt) = 15.0 m/s

For crate:-

Mass (m) = 12 kg

Initial velocity (u) = 0

displacement (d) = 2.4 m

Final velocity = 1.5 m/s

According to work done-kinetic energy theorem

The work done is equal to the change in the kinetic energy:

K = Kf Ki = W

W = (1/2)mv^2 - (1/2)mu^2

W = Force * displacement = (½) * 12 Kg * (1.5 m/s) ^2 – 0

Force * 2.4 m = 13.5

Force = 5.625 N

According to impulse-momentum theorem the change in momentum of an object equals the impulse applied to it.

Impulse is a vector quantity and has the same direction as the average force.

J = p

F t = m v =mv – mu

t = [ 12 Kg * (1.5 m/s) ] / 5.625 N

t = 3.2 sec

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