Question

a train is travelling up an a 3.73 degrees incline at a speed of 3.25m/s when the last car breaks free and begins to coast without friction.
How long does it take for the last car to come to rest momentarily?

any idea how to approach this question? is it just using the equations of motion with sin and cos?

Answer 1

Do you have the mass of the car? if yes it is simple then, you can use the equation:
\[V = at+V _{0}\] You know that \[V _{0}=3.25 m/s\] and V is equal to zero (it stops at the end) so you only need to find the acceleration a (which must be negative) using Newton's 2nd law: \[-mgsin \alpha=ma\]then\[a=-g \sin \alpha\](and \[\alpha=3.73\]
Is this what you are looking for?

Answer 2

i dont have the mass. if i find the distance by constructing a triangle with a right angle and the angle of 3.73 degrees and finding the length of the hypotneous to use as the distance but it comes out as 0 when multiplied by the 0 acceleration.....?

Answer 3

You really dont need the mass of the car. The component of acceleration in the direction of velocity is good enough. It is the same for any mass. And that would indeed be g sin(a). This g sin(a) will decelerate the car to rest.

The key is to understand how the weight of the car is distributed into components. There is one component that is perpendicular to the incline = mg cos(a). But this will get nullified by the normal reaction from the incline (any rigid surface produces a normal reaction that is perpendicular to the surface and opposite to force applied on it). Thus this component is inconsequential to our calculation.

The other force mg sin(a) will be in the opposite direction of velocity of the car. This directly decelerates the car.

Answer 4

The distance is found from equation: \[v ^{2} = u ^{2} + 2aS\]

For this problem, the solution (distance to rest) turns out as: \[D = u ^{2} / (2g \sin(\alpha))\]
The solution is: 8.28m

Answer 5

thank you