How Much Force Did the Rocket Exert on the Sled?

[freshcoast]

1. Problem statement, all variables and given/known data.

A sled weighs 2 Newtons. It is set in motion over a frictionless surface by a rocket motor with a weight of 1 Newton. After the fuel has been expended, the sled is coasting at a speed of 2m/s. How much horizontal force did the rocket exert on the sled while the fuel was burning? You may ignore the mass of the fuel.

Is it a) 4N
b) 6N
c) 12N
d) Not enough information

2. Homework Equations .
Constant acceleration
Vf = vo + at
X = volt + 1/2at^2
Vf^2 = vo^2 + 2ax

F = ma

3. Attempt.
I drew a free body diagram but it lead me no where cause I don't see how I can use any of the vertical components to the find the horizontal force.

 

[Simon Bridge]

Since the sled+rocket is moving at a constant speed - what is the net force on it?
What did the rocket force do the the motion of the sled? OR - what is the equation for force? Compare the quantities in that equation with the quantities you are given.

 

[freshcoast]

Since the sled+rocket is moving at a constant speed - what is the net force on it?
What did the rocket force do the the motion of the sled? OR - what is the equation for force? Compare the quantities in that equation with the quantities you are given.

I am still having trouble understanding this.. so the sled and rocket moving at a constant speed would mean there is no acceleration so the net force would be 0. The rocket force caused the sled to move in the +I direction while exerting a force in the -I direction and the equation for force is F = ma..but I still don't understand how to tie in the final velocity and the weights..

 

[Simon Bridge]

Why do you think you need to tie in the final velocity with the weights?

Since there is no net force on the sled - it follows that the speed of the sled is not related to the weights in any diagram you can draw for the constant motion.
Since the rocket is fixed to the sled - you only need one fbd for the acceleration phase.

recap:
The rocket accelerated the sled.
The equation is F=ma so that's the main equation you use since the question asks you for F.

List what you know.

 

[freshcoast]

Still having a hard time trying to grasp this, so far I can list that if I divide the weights by gravity that will give me the mass in kg.. and I think the final velocity ties in with the weight to maybe find the acceleration? and if I try to use one of the constant acceleration equations to solve for a I am missing a change of x in my equation..

 

[PeterO]

1. Problem statement, all variables and given/known data.

A sled weighs 2 Newtons. It is set in motion over a frictionless surface by a rocket motor with a weight of 1 Newton. After the fuel has been expended, the sled is coasting at a speed of 2m/s. How much horizontal force did the rocket exert on the sled while the fuel was burning? You may ignore the mass of the fuel.

Is it a) 4N
b) 6N
c) 12N
d) Not enough information

2. Homework Equations .
Constant acceleration
Vf = vo + at
X = volt + 1/2at^2
Vf^2 = vo^2 + 2ax

F = ma

3. Attempt.
I drew a free body diagram but it lead me no where cause I don't see how I can use any of the vertical components to the find the horizontal force.

Surely the rocket motor is attached to the sled?

You can find the final momentum of the sled+rocket; so Impulse - momentum equations would be the next step. One of those options should be the answer you get.

 

[Simon Bridge]

Still having a hard time trying to grasp this, so far I can list that if I divide the weights by gravity that will give me the mass in kg.. and I think the final velocity ties in with the weight to maybe find the acceleration? and if I try to use one of the constant acceleration equations to solve for a I am missing a change of x in my equation..

Presumably you can take the initial velocity to be zero since it says that the sled "is set in motion": so you have change in velocity. But acceleration is change in velocity over change in time - do you have change in time?

PeterO is suggesting that you try the impulse relation: how do you get from an impulse to a force?