Why is the work of the skier against resistance not negative?

[phospho]

A girl and her sledge have a combined mass of 40kg. She starts from rest and descends a slope which is inclined at 25 degrees to the horizontal. At the bottom of the slope the ground becomes horizontal for 15m before rising at 6 degrees to the horizontal. The girl travels 25m up the slope before coming to rest once more. There is a constant resistance to motion of magnitude 18N. Calculate the distance the girl travels down the slope.

Let point A be the where she is at rest (the start)
let point B be when she is at the end of the slope, let her speed be v ms^-1
let point C be when she is about to go up the second slope, and let her speed here be u ms^-1
let point D be when she comes to rest again.

let x be the distance of the first slope

take g to be 9.8 ms^-2

from points AB:

Gain in KE: $0.5\times v^2 \times 40 = 20v^2$

Loss in GPE: $xsin25 \times 9.8 \times 40$

therefore by the work energy principle:

$392xsin(25) - 20v^2 = 18x$
$x(392sin(25) - 18) = 20v^2$

from points BC:

Loss in KE: $0.5\times u^2 \times 40 - 0.5 \times v^2 \times 40 = 20u^2 - 20v^2$

as there is no change in GPE, by the work energy principle: $20u^2 - 20v^2 = 15\times 18$
$20u^2 = 270 + 20v^2$

from points CD:

Loss in KE: $20u^2 - 0 = 20u^2$

gain in GPE: $25sin(6) \times 9.8 \times 40 = 9800sin(6)$

by the work energy principle:

$9800sin(6) - 20u^2 = 25\times 18$
$9800sin(6) - 450 - (270 + 20v^2) = 0$
$\frac{9800sin(6) - 720}{392sin(25) - 18} = x = 1.92...$

however this is wrong and the answer is 11.8m

help :(?

 

[ehild]

A girl and her sledge have a combined mass of 40kg. She starts from rest and descends a slope which is inclined at 25 degrees to the horizontal. At the bottom of the slope the ground becomes horizontal for 15m before rising at 6 degrees to the horizontal. The girl travels 25m up the slope before coming to rest once more. There is a constant resistance to motion of magnitude 18N. Calculate the distance the girl travels down the slope.

Let point A be the where she is at rest (the start)
let point B be when she is at the end of the slope, let her speed be v ms^-1
let point C be when she is about to go up the second slope, and let her speed here be u ms^-1
let point D be when she comes to rest again.

let x be the distance of the first slope

take g to be 9.8 ms^-2

from points AB:

Gain in KE: $0.5\times v^2 \times 40 = 20v^2$

Loss in GPE: $xsin25 \times 9.8 \times 40$

therefore by the work energy principle:

$392xsin(25) - 20v^2 = 18x$
$x(392sin(25) - 18) = 20v^2$

from points BC:

Loss in KE: $0.5\times u^2 \times 40 - 0.5 \times v^2 \times 40 = 20u^2 - 20v^2$

as there is no change in GPE, by the work energy principle: $20u^2 - 20v^2 = 15\times 18$
[itex] 20u^2 = -270 + 20v^2 [/itex]

from points CD:

Loss in KE: $20u^2 - 0 = 20u^2$

gain in GPE: $25sin(6) \times 9.8 \times 40 = 9800sin(6)$

by the work energy principle:

$9800sin(6) - 20u^2 = 25\times 18$
$9800sin(6) - 450 - (270 + 20v^2) = 0$
$\frac{9800sin(6) - 720}{392sin(25) - 18} = x = 1.92...$[/COLOR]

however this is wrong and the answer is 11.8m

help :(?

The work of the resistive force is negative. Check the red lines.

No need to calculate change of KE during the stages of the motion: it is enough to apply work-energy theorem for the initial and final positions of the girl.


ehild

 

[phospho]

The work of the resistive force is negative. Check the red lines.

No need to calculate change of KE during the stages of the motion: it is enough to apply work-energy theorem for the initial and final positions of the girl.


ehild

Hi, thanks for replying

How do you know the work of the resistive force is negative? Also, for every other question I had to calculate the change of KE during each stage - How do you know I can do it by just considering GPE?

 

[ehild]

Hi, thanks for replying

How do you know the work of the resistive force is negative? Also, for every other question I had to calculate the change of KE during each stage - How do you know I can do it by just considering GPE?

Does a resistive force increase or decrease speed?

The Work-Energy Theorem states that the change of KE during the motion of a body is equal to the work of all forces exerted on it. If a body moves from A to B and then from B to C it moves from A to C doesn't it? If the KE is zero both at A and C, it does not matter what it was in between. Gravity is a conservative force, its work depends only on the height difference between the initial and final points.

ehild

 

[phospho]

Does a resistive force increase or decrease speed?

The Work-Energy Theorem states that the change of KE during the motion of a body is equal to the work of all forces exerted on it. If a body moves from A to B and then from B to C it moves from A to C doesn't it? If the KE is zero both at A and C, it does not matter what it was in between. Gravity is a conservative force, its work depends only on the height difference between the initial and final points.

ehild

Right I get why you only consider GPE now.

So the work done against the resistive force is positive when it goes down the first slope because there is a gain in KE and negative elsewhere as there is a loss in KE? Is this always the case?

 

[ehild]

So the work done against the resistive force is positive when it goes down the first slope because there is a gain in KE and negative elsewhere as there is a loss in KE? Is this always the case?

The work of what? There are only two forces, gravity and the resistive force. Both do work. Remember: work=force times displacement.

Gravity does positive work when the sledge descends along the first slope, and negative work when the sledge rises along the second slope.

The work of the resistive force is negative during the whole motion.

ehild

 

[azizlwl]

Hope this article will help you.
http://en.wikipedia.org/wiki/Conservative_force

 

[phospho]

The work of what? There are only two forces, gravity and the resistive force. Both do work. Remember: work=force times displacement.

Gravity does positive work when the sledge descends along the first slope, and negative work when the sledge rises along the second slope.

The work of the resistive force is negative during the whole motion.

ehild

I still don't understand why it's negative:

http://www.examsolutions.net/maths-revision/mechanics/work-energy-power/work-energy-principle/undulating-slope/tutorial-1.php

Why is the work against resistance here not negative :\

 

[azizlwl]

Let point A be the where she is at rest (the start)
let point B be when she is at the end of the slope, let her speed be v ms^-1
let point C be when she is about to go up the second slope, and let her speed here be u ms^-1
let point D be when she comes to rest again.
----------------------------------------------
Since start and end with zero velocity, you should take out KE into consideration.
In between energy taken away is by friction, Fx.
Who supply the energy? Must be mother earth.
Here is the reason why you should know about "conservative force".

 

[ehild]

I still don't understand why it's negative:

http://www.examsolutions.net/maths-revision/mechanics/work-energy-power/work-energy-principle/undulating-slope/tutorial-1.php

Why is the work against resistance here not negative :\

The work of the skier AGAINST resistance means a positive work on himself . The work of the resistive force on the skier is negative.
The skier does work, by using his sticks and pushing himself with them.
The girl on the sledge does not do work on herself.

Remember, the work of a force F along a displacement Δd is W=FΔd cos(θ): the magnitude of the force multiplied by the magnitude of the displacement and the cosine of the angle between them. A resistive force acts against displacement so the angle between them is 180°, the cosine is -1...

ehild