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# M2 Work Energy Principle Question

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1. The question in question is this (q2 pg 161 of the edexcel M2 textbook)

A box of mass 2kg is projected down a rough inclined plane with speed 4ms-1. The plane is inclined to the horizontal at an angle b, where tanb = 4/3, and the coefficient of friction between the box and the plane is 0.5. The box is modelled as a particle and air resistance is negligible. Using the work-energy principle, find the speed of the box after it has moved a distance of 5m down the plane.

So this is what I did first.

Resolving perpendicular to the plane, R = 2g cosb = 2g * 3/5 = 6g/5
Resolving parallel to the plane, downwards positive, 2g * sinb - 0.5R = 2g * 4/5 - 3g/5 = g

So the net force acting down the plane is gN.

Ek before = 1/2 * 2 * 4^2 = 16J
Ek after = 1/2 * 2 * v^2 = v^2
So delta Ek = v^2 - 16

Ep before = 0
Ep after = 4 * 2g = 8g
So delta Ep = -8g

Using the work - energy principle,

Wd * s = delta Ek + delta Ep
g * 5 = v^2 - 16 -8g
13g = v^2 - 16 => v = 12.0 ms-1 (3sf)

I retried it without including the Ep

Wd * s = delta Ek
5g = v^2 - 16 => v = root 65 = 8.06 ms-1 (3sf)

8.06 ms-1 is the correct answer - so why is the potential energy not included in this question?
2. (Original post by TheRealInjustice)
The question in question is this (q2 pg 161 of the edexcel M2 textbook)

A box of mass 2kg is projected down a rough inclined plane with speed 4ms-1. The plane is inclined to the horizontal at an angle b, where tanb = 4/3, and the coefficient of friction between the box and the plane is 0.5. The box is modelled as a particle and air resistance is negligible. Using the work-energy principle, find the speed of the box after it has moved a distance of 5m down the plane.

So this is what I did first.

Resolving perpendicular to the plane, R = 2g cosb = 2g * 3/5 = 6g/5
Resolving parallel to the plane, downwards positive, 2g * sinb - 0.5R = 2g * 4/5 - 3g/5 = g

So the net force acting down the plane is gN.

Ek before = 1/2 * 2 * 4^2 = 16J
Ek after = 1/2 * 2 * v^2 = v^2
So delta Ek = v^2 - 16

Ep before = 0
Ep after = 4 * 2g = 8g
So delta Ep = -8g

Using the work - energy principle,

Wd * s = delta Ek + delta Ep
g * 5 = v^2 - 16 -8g
13g = v^2 - 16 => v = 12.0 ms-1 (3sf)

I retried it without including the Ep

Wd * s = delta Ek
5g = v^2 - 16 => v = root 65 = 8.06 ms-1 (3sf)

8.06 ms-1 is the correct answer - so why is the potential energy not included in this question?
Hello there,

If I understand you correctly, you are wondering why you do not need to consider the change in gravitational potential energy. You have already resolved the weight of the particle into its parallel and perpendicular components. As weight is the force that results from the presence of a gravitational field, the work done by the weight represents the change in the gravitational potential energy. You therefore do not need to include this along with the method of finding the resultant force (you would essentially be counting it twice).

Of course, you could tackle the question by considering the change in gravitational potential energy. This would involve you finding the the change in height above the ground and then multiplying it by the mass and gravitational field strength to obtain the work done by the weight. To find the change in kinetic energy, you would have to subtract the work done by friction from the change in gravitational potential. You could then add this to the original kinetic energy and then, of course, solve for v.

I hope that this has been helpful.

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Updated: June 12, 2016
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