Box on ground, incline (with friction), ground: find Kinetic Energy

[questtosuccess]

Homework Statement

A 15 kg block slides along a horizontal frictionless surface 3.0 m above the ground at a constant speed of 2.0 m/s. The block then slides down an incline that makes an angle of 35° with the horizontal and has a coefficient of kinetic friction equal to 0.30. After reaching the end of the incline, the box continues sliding horizontally across the frictionless ground. Calculate the kinetic energy of the crate as it slides

a. along the upper surface.

b. along the ground.

Homework Equations

for a) KE= 1/2 mv²

for b) E₁ = E₂ + Wnon-conservative

The Attempt at a Solution

for a) KE=1/2 mv² = 1/2 (15) (2)² = 30 J

for b) I'm not quite sure where to consider the box sliding the ramp is at the moment... at the top (where y = 3 m) or the bottom (where y = 0 m)... regardless, this is how I did it.. but I'm pretty sure it's wrong

I began by finding the speed of the box when it's traveling down the ramp. I compared the energy of the box traveling along the upper surface, and the energy of the box going down the ramp (considering it to be at y = 0 m).

E₁ = E₂ + Wnon-conservative

1/2mv₁² + mgy₁ = 1/2mv₂² + F_fdcos180

v₂ = \sqrt{} (2(mv₁²/2+ mgy₁ - \mu_k F_Nd (-1) )) /m

From here, I solved and found that v₂ = 8.79 m/s
I had a problem already though.. I wasn't sure what "d" was supposed to be in this case ... for \mu_k F_Nd cos180 ... I let d = 3m (is that right?)

Finally, I used the above info and compared the energy of the box going down the ramp (*** considering it to be at y = 3 m this time), and the energy of the box traveling along the lower horizontal surface.

E₂ + Wnon-conservative = E₃
mgy₂ + 1/2mv₂² + F_fdcos180 = 1/2mv₃²

Therefore,

KE₃ = mgy₂ + 1/2mv₂² + F_fdcos180

I solved and got KE₃ = 910 J

All of this seems very incorrect. I feel like I'm doing this all wrong. Please help ! Thank you.

 

[gneill]

Homework Statement

A 15 kg block slides along a horizontal frictionless surface 3.0 m above the ground at a constant speed of 2.0 m/s. The block then slides down an incline that makes an angle of 35° with the horizontal and has a coefficient of kinetic friction equal to 0.30. After reaching the end of the incline, the box continues sliding horizontally across the frictionless ground. Calculate the kinetic energy of the crate as it slides

a. along the upper surface.

b. along the ground.

Homework Equations

for a) KE= 1/2 mv²

for b) E₁ = E₂ + Wnon-conservative

The Attempt at a Solution

for a) KE=1/2 mv² = 1/2 (15) (2)² = 30 J

for b) I'm not quite sure where to consider the box sliding the ramp is at the moment... at the top (where y = 3 m) or the bottom (where y = 0 m)... regardless, this is how I did it.. but I'm pretty sure it's wrong

I began by finding the speed of the box when it's traveling down the ramp. I compared the energy of the box traveling along the upper surface, and the energy of the box going down the ramp (considering it to be at y = 0 m).

E₁ = E₂ + Wnon-conservative

1/2mv₁² + mgy₁ = 1/2mv₂² + F_fdcos180

v₂ = \sqrt{} (2(mv₁²/2+ mgy₁ - \mu_k F_Nd (-1) )) /m

From here, I solved and found that v₂ = 8.79 m/s
I had a problem already though.. I wasn't sure what "d" was supposed to be in this case ... for \mu_k F_Nd cos180 ... I let d = 3m (is that right?)

Finally, I used the above info and compared the energy of the box going down the ramp (*** considering it to be at y = 3 m this time), and the energy of the box traveling along the lower horizontal surface.

E₂ + Wnon-conservative = E₃
mgy₂ + 1/2mv₂² + F_fdcos180 = 1/2mv₃²

Therefore,

KE₃ = mgy₂ + 1/2mv₂² + F_fdcos180

I solved and got KE₃ = 910 J

All of this seems very incorrect. I feel like I'm doing this all wrong. Please help ! Thank you.

Hi questtosuccess, Welcome to Physics Forums.

You've got a good plan, using energy conservation methods to solve the problem. One thing though: you don't have to calculate velocity as an intermediate step if all you are looking for is the energy change.

Your initial value for the KE (for part (a)) is fine.

For the calculation of the energy change as the block slides down the ramp though, as you suspected you've got a bit of a problem with the friction force. What is the distance that the block moves in the direction of the friction force? (What surface does the friction force parallel?)

 

[haruspex]

as you suspected you've got a bit of a problem with the friction force. What is the distance that the block moves in the direction of the friction force? (What surface does the friction force parallel?)

Also, I don't see where you calculate F_f or F_N.

 

[questtosuccess]

Hi questtosuccess, Welcome to Physics Forums.

You've got a good plan, using energy conservation methods to solve the problem. One thing though: you don't have to calculate velocity as an intermediate step if all you are looking for is the energy change.

Your initial value for the KE (for part (a)) is fine.

For the calculation of the energy change as the block slides down the ramp though, as you suspected you've got a bit of a problem with the friction force. What is the distance that the block moves in the direction of the friction force? (What surface does the friction force parallel?)

Thank you, I'm glad to be apart of this forum :)

I found the velocity of the box on the incline (v₂) because I needed to plug it into find the energy of the box at situation 3 (where it's traveling along the ground at 0 m).
How else would I be able to solve if I hadn't done that??

The distance isn't given.. that's what's confusing. The surface that the friction force parallels is the incline surface ... I still don't get it.

HARUSPEX: I calculated Ff .. it's μFn ... and Fn= mgcos35 .. I then solved and got that answer.

*It would be greatly appreciated if someone could let me know whether my final answer was correct. Thank you.

 

[gneill]

Thank you, I'm glad to be apart of this forum :)

I found the velocity of the box on the incline (v₂) because I needed to plug it into find the energy of the box at situation 3 (where it's traveling along the ground at 0 m).
How else would I be able to solve if I hadn't done that??

Kinetic energy is a quantity in its own right. Yes, it's related to velocity and mass by the formula (1/2)mv², but you don't need to go from KE to velocity and back again if you have the relationship that relates all the changes in energies involved. You can work with the energy figures directly.

The distance isn't given.. that's what's confusing. The surface that the friction force parallels is the incline surface ... I still don't get it.

Well, with the given change in height and the known angle of the slope, how long is the slope?

 

[questtosuccess]

Kinetic energy is a quantity in its own right. Yes, it's related to velocity and mass by the formula (1/2)mv², but you don't need to go from KE to velocity and back again if you have the relationship that relates all the changes in energies involved. You can work with the energy figures directly.

Well, with the given change in height and the known angle of the slope, how long is the slope?

Do you mean I should just set the equation as E₁= E₂ + Wnon-conservative = E₃ ??

The slope is 35 degrees... and the change in height is 3m right? Because the top is 3m and the bottom is 0m ... ?

 

[gneill]

Do you mean I should just set the equation as E₁= E₂ + Wnon-conservative = E₃ ??

You start with some initial kinetic energy KE_i that you calculated (30 J) before. As the box falls in elevation, gravitational PE will be converted to more KE for the box. You know how to calculate that amount. Also as the box slides on the surface of the ramp it will convert some KE to heat due to the action of friction, so the box loses some energy there. You know that amount of lost energy due to the work done by friction is given by f_fd, where f_f is the frictional force and d the distance along the surface that the force acts.

KE_i + PE_i = KE_f + PE_f + f_fd

Solve for KE_f. No velocities are needed for this part since you're actually looking for KE_f and you calculated KE_i previously.

The slope is 35 degrees... and the change in height is 3m right? Because the top is 3m and the bottom is 0m ... ?

Draw the picture. You should see a triangle... What side of the triangle does the box slide on? How long is that side?

 

[siddharth23]

Thank you, I'm glad to be apart of this forum :)

I found the velocity of the box on the incline (v₂) because I needed to plug it into find the energy of the box at situation 3 (where it's traveling along the ground at 0 m).
How else would I be able to solve if I hadn't done that??

The distance isn't given.. that's what's confusing. The surface that the friction force parallels is the incline surface ... I still don't get it.

HARUSPEX: I calculated Ff .. it's μFn ... and Fn= mgcos35 .. I then solved and got that answer.

*It would be greatly appreciated if someone could let me know whether my final answer was correct. Thank you.

The final answer I got was 287.37 J. Is the answer given anywhere?

 

[questtosuccess]

You start with some initial kinetic energy KE_i that you calculated (30 J) before. As the box falls in elevation, gravitational PE will be converted to more KE for the box. You know how to calculate that amount. Also as the box slides on the surface of the ramp it will convert some KE to heat due to the action of friction, so the box loses some energy there. You know that amount of lost energy due to the work done by friction is given by f_fd, where f_f is the frictional force and d the distance along the surface that the force acts.

KE_i + PE_i = KE_f + PE_f + f_fd

Solve for KE_f. No velocities are needed for this part since you're actually looking for KE_f and you calculated KE_i previously.

Draw the picture. You should see a triangle... What side of the triangle does the box slide on? How long is that side?

I just got it :D ! Thank you so much ! it's just trig ratios, so d= 3/sin35
Your help was much appreciated !

 

[questtosuccess]

The final answer I got was 287.37 J. Is the answer given anywhere?

Nope, no answer. But I got 282.1 J... so I'm assuming there was just an error in the rounding off throughout calculations.

Thanks for the reply ! :)

 

[siddharth23]

Anytime mate :)