SamKeene
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So I can't wrap my head around this question, despite it obviously being so simple. I've learnt the 'work done' equation(s):

Work done (Fs) =
Fs=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

At an angle:
Fs\cos{\theta}=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

With resistance force:
Fs\cos{\theta}-Rs=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

The question is:

A bicycle of mass 30kg is pushed up a hill inclined at 15 degrees to the horizontal. Calculate the work done in moving the bicycle 70 metres, starting and finishing with the bicycle at rest.

This 'Work, Energy and Power' section just isn't clicking with me, can someone give me a lil push?
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SherlockHolmes
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(Original post by SamKeene)
So I can't wrap my head around this question, despite it obviously being so simple. I've learnt the 'work done' equation(s):

Work done (Fs) =
Fs=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

At an angle:
Fs\cos{\theta}=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

With resistance force:
Fs\cos{\theta}-Rs=\frac{1}{2}mv^{2}-\frac{1}{2}mu^{2}

The question is:

A bicycle of mass 30kg is pushed up a hill inclined at 15 degrees to the horizontal. Calculate the work done in moving the bicycle 70 metres, starting and finishing with the bicycle at rest.

This 'Work, Energy and Power' section just isn't clicking with me, can someone give me a lil push?
Start by drawing a diagram. Then resolve parallel to the plane to get the force acting down the plane. Use Work Done = Force x Distance .
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SamKeene
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(Original post by SherlockHolmes)
Start by drawing a diagram. Then resolve parallel to the plane to get the force acting down the plane. Use Work Done = Force x Distance .
Fs - Rs = 0

Fs=Rs

 R = \frac{30g}{\cos{75}} Where R is the hypotenuse of the triangle formed.

But that doesn't work, it should be 30g*cos(75), but why? When resolving up the plane, isn't the amount of resistance of 30g the hypotenuse of the triangle formed?
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SherlockHolmes
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(Original post by SamKeene)
Fs - Rs = 0

Fs=Rs

 R = \frac{30g}{\cos{75}} Where R is the hypotenuse of the triangle formed.

But that doesn't work, it should be 30g*cos(75), but why? When resolving up the plane, isn't the amount of resistance of 30g the hypotenuse of the triangle formed?
The only force acting down the plane is the parallel component of the weight.

If you resolve parallel to the plane, you get mgsin(15) which is equal to mgcos(75).

http://upload.wikimedia.org/wikipedi...e_body.svg.png
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SamKeene
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(Original post by SherlockHolmes)
The only force acting down the plane is the parallel component of the weight.

If you resolve parallel to the plane, you get mgsin(15) which is equal to mgcos(75).
But \sin{15}=\frac{30g}{R} where R is the slope.

R\sin{15}=30g

How come you have mg on the other side?
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SherlockHolmes
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(Original post by SamKeene)
But \sin{15}=\frac{30g}{R} where R is the slope.

R\sin{15}=30g

How come you have mg on the other side?
Your triangle should have mg as the hypotenuse because you are splitting it into its vertical and horizontal components.
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