A worker on a flatbed railroad car is pushing a crate. The train is moving at a

Question

A worker on a flatbed railroad car is pushing a crate. The train is moving at a constant speed of 15 m/s. The crate has a mass of 12 Kg, and in being pushed over a distance of 2.4 m. Its velocity has increased (relative to the car) at a constant acceleration from rest to 1.5 m/s. (a) What are the crate's displacement, intial velocity, final velocity, and acceleration as measured by an abserver inside the train and another standing on the ground? (b) Find the force acts on the crate as measure by each observer. (c) Show that the work done on the crate is equal to the change in its kinetic energy for each observer. (d) If the crate starts to move with constant velocity of 1.5 m/s at the end of the 2.4 m and the driver of the train appplies the brakes, the train begins to decelerate with constant rate of -2.8 m/s^2, find the force acts on the crate according to each observer?

Ok here is how far I have gotten:

Crate on train:

m= 12 kg

X(final)= 2.4 m

V=1.5 m/s

V(intial)=0 m/s

the displacement:

X(final)-X(intial)= 2.4 m

V^2=V(intial)^2 + 2a( Delta X)

a= 0.47 m/s

V=V(intial) + at

t=3.19 sec.

Outside the train:

V(intial)= 0+15= 15 m/s

V(final)= 1.5 + 15= 16.5

x=(V+Vo)t / 2

2.4=(1.5)t/2

t= 3.19

V(final)=V(intial) +at

16.5=15 + a(3.19)

a=0.47 m?s^2

x= Vot+.5at^2

x=15(3.19)+.5(.47)(3.19)^2

x= 50.24

Work done

In: W=Fd

W=mad

W=12(.47)(2.4)

W= 13.53 J

Out:

W=Fd

W=mad

W=12(.47)(50.24)

W=283.35 J

Thats How far I got. I'm not sure how to solve the rest. Any help would be fantastic! Thx! (P.S if you don't mind could show me how you got that answer)

Explanation / Answer

similar,

a) According to the work-energy theorem,
work done by the worker = KE gained by the crate
=> F*s = (1/2) m v2
=> 5.63*2.4 = (1/2)*12*v2
=> v2 = 5.63*2.4*2/12
=> v2 = 2.252
=> v ~ 1.5015 m/s.

b) The work done, W' = F' * s'
= 5.63 * (x+2.4) ---- (1)

To proceed we need x, the distance moved by the train in the timetaken by the worker to move the box 2.4 m, to proceed. This can befound as x = speed of the train * time to move the box.
=> x = 15 * sqrt(2*s/a) using the formula s = ut + (1/2) at2 with u = 0 since the crate is said to start fromrest.
=> x = 15*sqrt(2*2.4/{F/m}) (using F = m*a)
=> x = 15 * sqrt (4.8*12/ 5.63)
=> x ~ 47.9787 m

Thus, going back to (1), W' ~ 5.63*(47.9787+2.4) = 283.63 J

In the frame O' the crate has an initial velocity, u' = 15 m/s andaccelerates at F/m = 5.63/12 ~ 0.46917 m/s2 and travelsa distance of s' = x+2.4 ~ 47.9787+2.4 ~ 50.3787 m. Therefore,using v'2 - u'2 = 2*a'*s' => v'2 -u'2 =2*a*50.3787(a=a' is given)
= 2*0.46917*50.3787.

Therefore, the change in kinetic energy in this frame,
?KE' = final KE - initial KE =(1/2)*m*v'2 - (1/2)*m*u'2
=(1/2)*m*(v'2-u'2)
~(1/2)*12*(2*0.46917*50.3787)
~ 283.63 J
=> ?KE' = W'

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