Work and Energy of an electric motor

[Inertialforce]

Homework Statement

An electric motor and a rope are used to pull a 10kg crate of car parts up an inclined plane. The crate starts out from rest on the ground and ends up with a speed of "vf" at a height of 4.0m above the ground.

The graph provided shows the force exerted on the crate by the motor as it is pulled 10m up the inclined plane.

a) How much work is done on the crate by the electric motor from d=0 to d=10?

b) 150J of heat energy is produced through friction during the 10m pull. What is the final speed of the crate at d=10m?

Homework Equations

a) Work= area under the graph
Area of a Trapezoid = (1/2)*(a+c)*(b)

b) Δ E = Heat

The Attempt at a Solution

a) Work = area under the graph
W= (1/2)*(a+c)*(b)
W= (1/2)*(50+65)*(10)
W= 575J

b)ΔE= Heat
ΔEp + ΔEk = Heat
(Epf - Epi) + (Ekf - Eki) = 150J
(Epf- 0) + (Ekf - 0) = 150J
mghf + (1/2)mvf(squared) = 150J
((10)(9.80)(4.0)) + ((1/2)(m)(vf(squared)) = 150J
392 + ((1/2)(10)(vf(squared)) = 150J

this is where I run into the problem, if I try to isolate the vf by moving the 392 to the right side to subtract from the 150J I get a negative number which I will then have to square root which is impossible. I was just wondering whether I was going in the right direction ( for both questions "a" and "b") and if I am how am I supposed to isolate the vf here so that I can get an answer when I square root it?

 

[alphysicist]

Hi Inertialforce,

I can't see your attachment yet, but something does not seem right to me.

Homework Statement

An electric motor and a rope are used to pull a 10kg crate of car parts up an inclined plane. The crate starts out from rest on the ground and ends up with a speed of "vf" at a height of 4.0m above the ground.

The graph provided shows the force exerted on the crate by the motor as it is pulled 10m up the inclined plane.

a) How much work is done on the crate by the electric motor from d=0 to d=10?

b) 150J of heat energy is produced through friction during the 10m pull. What is the final speed of the crate at d=10m?

Homework Equations

a) Work= area under the graph
Area of a Trapezoid = (1/2)*(a+c)*(b)

b) Δ E = Heat

The Attempt at a Solution

a) Work = area under the graph
W= (1/2)*(a+c)*(b)
W= (1/2)*(50+65)*(10)
W= 575J

b)ΔE= Heat
ΔEp + ΔEk = Heat

These last two equations are not correct. There are two things wrong: the work you just found in part a is not included in your equation, and also you are not quite treating the thermal energy increase correctly.

Here is the work-energy equation

$$W_{\rm nc} = \Delta E$$
so that the work done by non-conservative forces equals the change in energy.

The frictional force does a certain amount of negative work on the crate that shows up as heat. At this point we have a choice in how we view the problem. We can either track the work done by the frictional force or the increase in thermal energy. In this case, you are tracking the heat energy, so the +150J needs to be on the energy side of the equation.

(If instead of the heat, you were tracking the frictional work, it would -150J on the work side of the equation. So mathematically there is no difference.)

 

[Inertialforce]

Hi Inertialforce,

I can't see your attachment yet, but something does not seem right to me.



These last two equations are not correct. There are two things wrong: the work you just found in part a is not included in your equation, and also you are not quite treating the thermal energy increase correctly.

Here is the work-energy equation

$$W_{\rm nc} = \Delta E$$
so that the work done by non-conservative forces equals the change in energy.

The frictional force does a certain amount of negative work on the crate that shows up as heat. At this point we have a choice in how we view the problem. We can either track the work done by the frictional force or the increase in thermal energy. In this case, you are tracking the heat energy, so the +150J needs to be on the energy side of the equation.

(If instead of the heat, you were tracking the frictional work, it would -150J on the work side of the equation. So mathematically there is no difference.)

Would this be more accurate (or correct)?:

Wnc = ΔE
Wnc = (Epf-Epi) + (Ekf-Eki) + heat
575 = (Epf) + (Ekf) + heat
575 = (mghf) + ((1/2)mvf(squared)) + 150
575 - 150 = (10)(9.80)(4.0) + (1/2)(10)(vf(squared))
425 = 392 + (1/2)(10)(vf(squared))
425 - 392 = (1/2)(10)(vf(squared))
(33)(2) = (10)(vf(squared))
66/10 = vf(squared)
√6.6 = √vf(squared)
2.569046516 = vf
2.6 m/s = vf

Sorry but could you check and tell me if this is correct or if I am doing it correctly now, because our class has never (or hasn't yet) learned this equation (Wnc = ΔE) so I am unsure as to how to utilize the formula.

 

[alphysicist]

Would this be more accurate (or correct)?:

Wnc = ΔE
Wnc = (Epf-Epi) + (Ekf-Eki) + heat
575 = (Epf) + (Ekf) + heat
575 = (mghf) + ((1/2)mvf(squared)) + 150
575 - 150 = (10)(9.80)(4.0) + (1/2)(10)(vf(squared))
425 = 392 + (1/2)(10)(vf(squared))
425 - 392 = (1/2)(10)(vf(squared))
(33)(2) = (10)(vf(squared))
66/10 = vf(squared)
√6.6 = √vf(squared)
2.569046516 = vf
2.6 m/s = vf

Sorry but could you check and tell me if this is correct or if I am doing it correctly now, because our class has never (or hasn't yet) learned this equation (Wnc = ΔE) so I am unsure as to how to utilize the formula.

The form of the equation looks right to me now; however I can't yet say everything is correct until the attachment in your original post is approved.