Just learning about work,energy

[CollegeStudent]

Homework Statement

Starting from rest 10.0 kg mass slides 5.00 m down a 20.0 incline. The coefficient of kinetic friction is 0.300. (a) Determine the work done by the force of gravity. (b) Determine the work done by the force of kinetic friction. (c) Determine the change in gravitational potential energy. (d) Determine the kinetic energy of the block at the bottom.

Homework Equations

umm...

kinetic energy = 1/2mv² i believe
gravitational potential energy = mass*gravity*height
Work = Force * distance * cosθ

The Attempt at a Solution

well for part A) I was thinking
(10.0kg*9.81m/s²) * 5.00m * cos(20)

for part B) I was thinking
.300 * 5.00m * cos(20)

for part C) I was thinking
10.0kg * 9.81m/s² * (5.00m *sin20)to find the vertical height??

and for part D) I was thinking it would be 0 because at the bottom...velocity would be 0...so 1/2*mass*0² = 0

are any of these assumptions correct?

 

[mfb]

Be careful with the angle at (a).
At b: You need the force of friction first. This is not the same as the coefficient of friction (which you can see from the mismatching units, for example).
Why do you think the velocity is 0 at the bottom?

 

[rude man]

Homework Statement

Starting from rest 10.0 kg mass slides 5.00 m down a 20.0 incline. The coefficient of kinetic friction is 0.300. (a) Determine the work done by the force of gravity. (b) Determine the work done by the force of kinetic friction. (c) Determine the change in gravitational potential energy. (d) Determine the kinetic energy of the block at the bottom.

Homework Equations

umm...

kinetic energy = 1/2mv² i believe
gravitational potential energy = mass*gravity*height
Work = Force * distance * cosθ

The Attempt at a Solution

well for part A) I was thinking
(10.0kg*9.81m/s²) * 5.00m * cos(20)

cos ?

for part B) I was thinking
.300 * 5.00m * cos(20)

No. what's the expression for work done by kinetic friction force thru a distance?

for part C) I was thinking
10.0kg * 9.81m/s² * (5.00m *sin20)to find the vertical height??

Check.

and for part D) I was thinking it would be 0 because at the bottom...velocity would be 0...so 1/2*mass*0² = 0

why do you assume v = 0 at the bottom?
Rather, use energy conservation: Δp.e. + Δk.e. + friction loss = 0.

are any of these assumptions correct?[/QUOTE]

see above

 

[CollegeStudent]

Be careful with the angle at (a).
At b: You need the force of friction first. This is not the same as the coefficient of friction (which you can see from the mismatching units, for example).
Why do you think the velocity is 0 at the bottom?

why won't the angle be 20 degrees in part (a)? It's not a straight down plummet...but is the fact that gravity acts straight down an indicator that the angle sould be 0 degrees?

and ohh okay So I'd have to solve the kinetic friction = μ_k * N first for part (b)

and I was thinking that the velocity would be 0 at the bottom because it doesn't specify where at the bottom...but just the bottom...so i figured that the kinetic (moving) energy would be the 1/2*10.0kg * velocity at the bottom, which I thought would be 0...but I suppose I am wrong?

 

[CollegeStudent]

For part (a) would I use *sin* instead of *cos* in the Work equation? since the block is moving in the same direction that mgsinθ is pointed in?


for (b)... we solve for friction

friction = μ_k * N

from the FBD...N = mgcosθ

so that work equation would look like

W = (.300 * mgcosθ) * 5.00m * cos(20)

is that correct? because that is the only "work" equation I see on my sheet

 

[rude man]

View attachment 57287

View attachment 57288

For part (a) would I use *sin* instead of *cos* in the Work equation? since the block is moving in the same direction that mgsinθ is pointed in?

Good move. The component of gravity in the direction of the ramp is mgsin(20).

for (b)... we solve for friction

friction = μ_k * N

from the FBD...N = mgcosθ

so that work equation would look like

W = (.300 * mgcosθ) * 5.00m * cos(20)

is that correct? because that is the only "work" equation I see on my sheet

Wy the cos(20) term at the end? Your force is 0.3mgcos(θ) and your distance is 5m. End of story.

 

[CollegeStudent]

Wy the cos(20) term at the end? Your force is 0.3mgcos(θ) and your distance is 5m. End of story.

OH WAIT! I remember my prof saying "Don't use the angle in the problem!"
Is it...the fact that since the force (friction) is acting in the opposite direction as the displacement...then the equation would be cos(180) a.k.a (1)...this is why you are telling me don't worry about the cos correct??

That's why it is just the force of friction times the displacement correct??

 

[CollegeStudent]

Good move. The component of gravity in the direction of the ramp is mgsin(20).

great! alright...And if my last statement is correct...the the fact that the force here is mgsin theta which creates a 90° angle with the displacement...then the cos would be of 90°...which would = 0

So then wouldn't the work done by the force of gravity = 0?

 

[rude man]

OH WAIT! I remember my prof saying "Don't use the angle in the problem!"
Is it...the fact that since the force (friction) is acting in the opposite direction as the displacement...then the equation would be cos(180) a.k.a (1)...this is why you are telling me don't worry about the cos correct??

That's why it is just the force of friction times the displacement correct??

I don't know why your prof said " ... don't use the angle in the problem ...". Thing is , you used it twice!

Look, the force between the block and the ramp is mg cosθ. So the force of kinetic friction is that times the kinetic friction coefficient. And the distance is 5m. Like I said, end of story. Work = force times displacement.

 

[rude man]

great! alright...And if my last statement is correct...the the fact that the force here is mgsin theta which creates a 90° angle with the displacement...then the cos would be of 90°...which would = 0

So then wouldn't the work done by the force of gravity = 0?

No, gravity work = force of gravity along the ramp times ramp distance.

 

[CollegeStudent]

I don't know why your prof said " ... don't use the angle in the problem ...". Thing is , you used it twice!

Look, the force between the block and the ramp is mg cosθ. So the force of kinetic friction is that times the kinetic friction coefficient. And the distance is 5m. Like I said, end of story. Work = force times displacement.

THANK YOU for saying it EXACTLY like that! that made SO much sense!

So freakin' obvious too!

so

(mgcosθ * .300) * 5.00m

No, gravity work = force of gravity along the ramp times ramp distance.

ohhh...

so then

mgsinθ * 5.00m

(NO cosθ at the end because it was already used! Again THANK YOU for pointing this out!)

 

[CollegeStudent]

For part B) again (the friction problem)...the force would be "negative" correct? since it is acting in the opposite direction?

 

[rude man]

For part B) again (the friction problem)...the force would be "negative" correct? since it is acting in the opposite direction?

Yes, F_f opposes the direction of displacement so F_f*d < 0. But what really matters is there is energy expended (lost) so the k.e. at the end will be the p.e. at the beginning minus the friction energy lost.

BTW work and energy are the same thing.

Got part (d) yet?

 

[CollegeStudent]

hmm part (d)

well from what I read about what you said (conservation of energy) it looks like I would take

the gravitational potential energy and add the friction energy (the 2 answers I got)

so would it be

(mgsinθ * 5.00m) + (-mgcosθ * .300) * 5.00m) = Kinetic Energy?

is that correct??

 

[rude man]

hmm part (d)

well from what I read about what you said (conservation of energy) it looks like I would take

the gravitational potential energy and add the friction energy (the 2 answers I got)

so would it be

(mgsinθ * 5.00m) + (-mgcosθ * .300) * 5.00m) = Kinetic Energy?

is that correct??

It actually is but you should think as follows:

potential energy loss = mgh where h = 5*sinθ in this case
friction energy lost = 5*0.3*mgcosθ
therefore kinetic energy gained = potential energy lost - friction energy loss

Take note that "p.e. lost" and "friction loss" are both positive numbers the way I've defined them.
In other words, you've lost p.e. so you should have gained that much k.e. but some of the k.e. is discounted due to friction loss along the way.

(Like in money: you lose $100, you don't lose -$100. Etc.)