# Homework Help: Force, velocity, distance, direction

1. Jul 30, 2009

### dance_sg

1. The problem statement, all variables and given/known data
A barrel of oil with a mass of 65 kg is rolled down a 1.2 m ramp. Due to a constant force exerted on it, the barrel reaches a velocity of 4.2 m/s just on leaving the ramp. Find the constant force on the barrel while it is on the ramp and express your answer in the form of a sentence relating force, change in velocity, distance, and direction.

2. Relevant equations

f=ma, v=d/t, a=v/t,

3. The attempt at a solution
i have no idea how to even start this question. should i find the time to plug it into the a=v/t formula, then use that to put it into the f=ma formula? how do i find change in velocity, distance and direction??

2. Jul 30, 2009

### kuruman

Re: force,velocity,distance,direction

Careful here. If there is rolling, the use of rotational kinematics and dynamics is in order. The point on the barrel where the constant force is applied becomes important because torques are be involved. Where is that point?

3. Jul 30, 2009

### dance_sg

Re: force,velocity,distance,direction

just as it is leaving the ramp?

4. Jul 30, 2009

### kuruman

Re: force,velocity,distance,direction

If it is a constant force, it must be applied on the barrel at all times while the barrel rolls down. Otherwise the force would not be constant.

5. Jul 30, 2009

### dance_sg

Re: force,velocity,distance,direction

oo. ok. so then i would use the formula a=vf^2-vi^s/2x?? then what ever i get for that mulitiply by the mass to find force??

6. Jul 30, 2009

### kuruman

Re: force,velocity,distance,direction

This will not work. Mass times acceleration is the net force acting on an object, that is the sum of all the forces. Here there are additional forces acting on the barrel, i.e. gravity and the ramp contact force.

Have you listed all that is given in the problem? For example, what is the slope of the ramp? It seems to me that the force required will be larger as the ramp becomes steeper because the speed at the end of the ramp is fixed. Also, the required force will be smaller if it is applied on the rim of the barrel as opposed to on its axis. That is why the point of application of the force is important.

7. Jul 30, 2009

### dance_sg

Re: force,velocity,distance,direction

ok, so how do i find out the force then?? was finding the acceleration the way i did correct??

and yes, i listed all that was given to me in the problem.

8. Jul 30, 2009

### kuruman

Re: force,velocity,distance,direction

Your method for finding the acceleration is correct. If nothing more is given, then we have to assume that the ramp is horizontal. In that case, the force exerted by the ramp on the barrel is in the vertical direction and cancels the force of gravity. This means that the net force is in the horizontal direction and is exactly the force we are looking for. So multiplying the acceleration by the mass as you initially suggested will do it.

Sorry about the confusion. The problem would have been easier to understand if "horizontal" were added before the word "ramp".

9. Jul 30, 2009

### dance_sg

Re: force,velocity,distance,direction

ok :)
so for the part about the sentence, i need to say the change in velocity from the top to the bottom of the ramp, would there be no change??? and the distance covered by the barrel would just be the 1.2m??

10. Jul 30, 2009

### kuruman

Re: force,velocity,distance,direction

If the acceleration is not zero, the velocity must change. Yes, the distance covered by the barrel is 1.2 m.

11. Jul 30, 2009

### dance_sg

Re: force,velocity,distance,direction

ok thank you :)