# Block sliding on an inclined plane

#### JoshMP

1. Homework Statement

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30 ^\circ angle. The block's initial speed is 10\;{\rm m}/{\rm s}. The coefficient of kinetic friction of wood on wood is \mu_k=.200.

What speed does it have when it slides back down to its starting point?

2. Homework Equations

F=ma
v^2= vi^2 +2as

3. The Attempt at a Solution

I've found the net force along the x-axis, used that to find the acceleration along x. I used this acceleration in the kinematic equations but I keep getting the wrong answer. I'm positive I found the correct acceleration because I got the correct answer for the first part of the question, which asked how far the block travels up the incline. The answer is NOT 10 m/s...(which I completely do not understand why it isn't).

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#### efekwulsemmay

Draw a free body diagram and all of the relevant forces.

The answer is NOT 10 m/s...(which I completely do not understand why it isn't).
Friction is constantly opposing the motion and slowing down your system. You have to take that into effect.

I might try a=F(net)\m for your kinematics equations. Remember for F(net) to include friction force.

#### JoshMP

Draw a free body diagram and all of the relevant forces.

Friction is constantly opposing the motion and slowing down your system. You have to take that into effect.

I might try a=F(net)\m for your kinematics equations. Remember for F(net) to include friction force.
I already did all that. I found Fnet to =13.2 N, which means a= -6.6 m/s^2. Since there is a constant acceleration (because there is a constant force, as you noted), I should be able to plug this value into the kinematic equations. I'm still getting the wrong answer.

#### efekwulsemmay

Honestly, you got me. Are you taking the 30 degree angle into equation? They gave it for a reason although right now I can't for the life of me figure it out.

#### JoshMP

Yeah, I used the angle when calculating the net force. Fnet= (friction) + (x component of gravity)= (mu*n) + (mgsin30)= (mu*mgcos30)+(mgsin30)

The force has to be found first. Using the force you can caluclate the acceleration, which must be constant. The acceleration allows you to use the kinematic equations.

#### efekwulsemmay

Wait, the force isn't constant. The friction force is constant. The net force is not. This is because the friction force is lowering the magnitude of the net force as the motion proceeds within the system. I am not sure how to express this mathematically, possibly
$$F_{net}=F_{final}-F_{initial}$$
or something to that effect. Don't take my word for it though. That formula is just a shot in the dark.

EDIT: also shouldn't the x component of gravity be mgcos30 and the normal mu*mgsin30?

#### w3390

I don't mean to interrupt you two working this out, but I am going to try to add some helpful input. I worked the problem and found the acceleration of the block moving up the ramp to be a=-6.6 m/s as well. I then used this number to find the distance the block traveled. Then when the block moves back down the ramp, you have to draw a new free-body diagram. This time the friction force and the force pulling the block down the ramp are not in the same direction the the acceleration is different. Using this new acceleration and the distance traveled in the first part, use the equation Vf^2-Vi^2=2*a*deltaX. This will give you a final velocity less than 10 m/s which I believe is correct.

#### JoshMP

I don't mean to interrupt you two working this out, but I am going to try to add some helpful input. I worked the problem and found the acceleration of the block moving up the ramp to be a=-6.6 m/s as well. I then used this number to find the distance the block traveled. Then when the block moves back down the ramp, you have to draw a new free-body diagram. This time the friction force and the force pulling the block down the ramp are not in the same direction the the acceleration is different. Using this new acceleration and the distance traveled in the first part, use the equation Vf^2-Vi^2=2*a*deltaX. This will give you a final velocity less than 10 m/s which I believe is correct.
Yep, that's it. I didn't realize I had to draw a new free body diagram. Thanks a lot both of you!