Solving Frictionless Plane Incline Problem: A & B

[Thoven]

A frictionless plane is 10.0 m long and inclined at 35.0°. A sled starts at the bottom with an initial speed of 5.00 m/s up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the incline with an initial speed vi. Both sleds reach the bottom of the incline at the same moment.

(a) Determine the distance that the first sled traveled up the incline.

(b) Determine the initial speed of the second sled.

I would give you an equation I had a clue of what to do if someone can lead me in the right direction with an equation or something I just don't know what to do at all.

 

[LowlyPion]

A frictionless plane is 10.0 m long and inclined at 35.0°. A sled starts at the bottom with an initial speed of 5.00 m/s up the incline. When the sled reaches the point at which it momentarily stops, a second sled is released from the top of the incline with an initial speed vi. Both sleds reach the bottom of the incline at the same moment.

(a) Determine the distance that the first sled traveled up the incline.

(b) Determine the initial speed of the second sled.

I would give you an equation I had a clue of what to do if someone can lead me in the right direction with an equation or something I just don't know what to do at all.

Welcome to PF.

First of all you need to calculate the acceleration component acting along the incline. This is your gravitational field.

Now you can calculate how long for the first sled to go up and stop. (This will be the same time for it to go back down again.) And also how far up it goes.

Knowing that you know how long the second sled has to reach the bottom and hence what speed it must be released at.

If you need equations look here:
https://www.physicsforums.com/showpost.php?p=905663&postcount=2

 

[Thoven]

Thank you very much that helped a lot I have another question now.

A 1000 kg car is pulling a 275 kg trailer. Together, the car and trailer have an acceleration of 2.28 m/s2 in the forward direction. Neglecting frictional forces on the trailer, determine the following (including sign).
(a) the net force on the car

(b) the net force on the trailer

(c) the force exerted by the trailer on the car

(d) the resultant force exerted by the car on the road

direction ° measured from the left of vertically downwards

I have the answer to A B and C they were easy just can't get D.
A=2280N
B=627N
C=-627N

 

[Thoven]

Nvm I have figured out the problem.