What is the displacement of a 3.0 kg block sliding down an inclined plane?

[Austin Gibson]

Homework Statement

:
A 3.0 kg block starts with a speed of 18 m/s at the bottom of a plane inclined at 41° to the horizontal. The coefficient of sliding friction between the block and plane is
μk = 0.22.
a) Use the work–energy principle to determine how far (in m) the block slides along the plane before momentarily coming to rest.

b) After stopping, the block slides back down the plane. What is its speed (in m/s) when it reaches the bottom? (Hint: For the round trip, only the force of friction does work on the block.)[/B]

Homework Equations

: 0.5m(v(final)^2 - v(initial)^2) = total work, Fcos(theta)d = work[/B]

The Attempt at a Solution

:

I sense that I'm misinterpreting the question.
[/B]

 

[Arman777]

I think you are doing wrong in the second block. Why did you used the force as mgsin41-friction to calculate the total work done ? Think about the directions of the forces.

You can also try to think from this equation
$TE_{initial}-TE_{final}=W_{friction}$

 

[Austin Gibson]

I think you are doing wrong in the second block. Why did you used the force as mgsin41-friction to calculate the total work done ? Think about the directions of the forces.

You can also try to think from this equation
$TE_{initial}-TE_{final}=W_{friction}$

I misinterpreted the question. If gravity and friction are resisting its movement, both would be negative. Therefore, the sum of the forces is -24.2N. When divided by the total work (-486 J), my answer is 20.1m.

 

[neilparker62]

In your work energy equations Fcos(θ) d become Fsin(θ) d because the angle between weight and slope is the complement of the slope angle. But to apply work energy in this case you need to compute F(net) d where F(net) = F sin(θ) - f(friction). Trust you are able to determine f(friction) (?)

To determine d you can just use one of the standard equations of motion noting that the component of g parallel to the slope is g sin(θ).

If your teacher is pedantic you might have to write F(net) = ∑F(parallel) = Fsin(θ) cos(0) + F(friction) cos (180). Endless confusion in these type of questions at school level is caused by there being (effectively) two "thetas" involved in the equations. One is the slope angle and the other is either zero or 180 degrees depending on whether the force component acts in the direction of motion or opposite.

Perhaps some careful thought should be given to standardise the way this is taught at schools.

 

[Austin Gibson]

In your work energy equations Fcos(θ) d become Fsin(θ) d because the angle between weight and slope is the complement of the slope angle. But to apply work energy in this case you need to compute F(net) d where F(net) = F sin(θ) - f(friction). Trust you are able to determine f(friction) (?)

To determine d you can just use one of the standard equations of motion noting that the component of g parallel to the slope is g sin(θ).

If your teacher is pedantic you might have to write F(net) = ∑F(parallel) = Fsin(θ) cos(0) + F(friction) cos (180). Endless confusion in these type of questions at school level is caused by there being (effectively) two "thetas" involved in the equations. One is the slope angle and the other is either zero or 180 degrees depending on whether the force component acts in the direction of motion or opposite.

Perhaps some careful thought should be given to standardise the way this is taught at schools.

The work done by friction by my calculations for the first question is -4.88d because I'm firstly solving for displacement. I summed the forces in the parallel direction first and then multiplied that number by the displacement (unknown). For the first question, I presume that the work done by gravity and friction will be negative because they're opposing the motion in the opposite direction.

 

[Austin Gibson]

Updated picture of my calculations:

 

[neilparker62]

Apologies I hadn't read your original post properly - that's all good and Arman777 picked the error in your workings correctly. What I wrote should be applicable to when the block slides down again.

 

[Austin Gibson]

Apologies I hadn't read your original post properly - that's all good and Arman777 picked the error in your workings correctly. What I wrote should be applicable to when the block slides down again.

Are you agreeing with my displacement value for the first question? I'd submit it, but my amount of submissions are low. Thus, I'm risking losing credit.

 

[Austin Gibson]

The following are my calculations for the second question presuming the displacement is 20.1:

 

[neilparker62]

Should be good. Please ignore what I said about using an equation of motion to find d since you used work/energy as required by the question.

 

[Austin Gibson]

Should be good. Please ignore what I said about using an equation of motion to find d since you used work/energy as required by the question.

I appreciate your assistance! Both answers were correct.

 

[neilparker62]

Great - although to be quite honest you had it all sorted except for the correction pointed out by Arman777!

 

[Austin Gibson]

Great - although to be quite honest you had it all sorted except for the correction pointed out by Arman777!

True, but I was slow to notice my error.