Percent of initial total mechanical energy lost during a round trip?

[bmx_Freestyle]

Given:
Mass= 5 kg
Vo= 40 m/s
Vf=50 m/s
s=150 m
Ho=0
Hf=50.2 m
g=9.8 m/s/s
friction=6.26 N
θ=30°

*You will probably not use all of these values*
What percent of initial total mechanical energy is lost during the round trip? (a block (with mass m) going up an inclined plane, and then back down)


i have no clue what to do...i think in end youll multiply something by 100% to get a percent? i honestly don't know

 

[PeterO]

Given:
Mass= 5 kg
Vo= 40 m/s
Vf=50 m/s
s=150 m
Ho=0
Hf=50.2 m
g=9.8 m/s/s
friction=6.26 N
θ=30°

*You will probably not use all of these values*
What percent of initial total mechanical energy is lost during the round trip? (a block (with mass m) going up an inclined plane, and then back down)


i have no clue what to do...i think in end youll multiply something by 100% to get a percent? i honestly don't know

Not too sure where these apply. [variables I highlighted red]

subscript o usually means initial, but is less than subscript f - the final.

not sure what the H values refer to?

The round trip is what?

 

[bmx_Freestyle]

Not too sure where these apply. [variables I highlighted red]

subscript o usually means initial, but is less than subscript f - the final.

not sure what the H values refer to?

The round trip is what?

H refers to height
i drew a right triangle and the vertical leg represents height
the bottom is Ho and the top of the vertical leg is Hf

the round trip is when the mass moves up the inclined plane and comes back down

 

[bmx_Freestyle]

Vo (40 m/s) is the speed the mass begins with when moving up the plane. when it stops the vf is o.
then when the mass comes back down vo=o and the vf is 50 (when the mass reaches the bottom)

 

[PeterO]

H refers to height
i drew a right triangle and the vertical leg represents height
the bottom is Ho and the top of the vertical leg is Hf

the round trip is when the mass moves up the inclined plane and comes back down

I just came across your earlier post.

As the mass moves up the slope, the initial kinetic energy will transform into some potential energy with some mechanical energy lost overcoming friction.
The 30^o slope means that for every metre you rise, you travel 2m along the slope.

I would be solving this as linear motion. Resolve the weight force into a component parallel to the slope - that will slow it down - and perpendicular to the slope - which will enable you to calculate the friction - an additional force slowing it down.

for example:
Suppose the parallel component was 20N, and the friction force was 5N [neither of those figures is correct, it is just for example]

On the way up the slope, that would mean a total retarding force of 25N. [20 + 5]
Given a mass of 5 kg, the acceleration would be 5 ms^-2.

You can use motion equations to find where the mass stops.

On the way back down, friction will be acting up the slope [opposing motion] so the net force is 15N [20 - 5]

SO the acceleration is only 3 ms^-2.

You can use motion equations to find the final speed.

by calculating KE at the start, and the kinetic energy at the end you can calculate the loss, and express it as a percentage.

All you have to do now is calculate the actual figures.

 

[PeterO]

Vo (40 m/s) is the speed the mass begins with when moving up the plane. when it stops the vf is o.
then when the mass comes back down vo=o and the vf is 50 (when the mass reaches the bottom)

You can be certain that the mass will NOT get to the bottom at a speed faster than it left!

 

[bmx_Freestyle]

Srry, but do u mean potential energy transforming into kinetic energy?

 

[bmx_Freestyle]

ok i calculated initial KE and final KE
KEo=1/2mvo^2
KEf=1/2mvf^2

KEo=1/2(5)(40^2)=4000
KEF=1/2(5)(50^2)=6250
then i did 4000/6250 and got 0.64 and then i multiplied tht by 100% to get 64%