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need a memory boost (kinematics/particle on a slope question) M2 watch

1. A lorry of mass 2000 kg is moving down a straight road inclined at angle a to the horizontal, where sina =1/25 . The resistance to motion is modelled as a constant force of magnitude 1600 N. The lorry is moving at a constant speed of 14 m s–1.

Find, in kW, the rate at which the lorry’s engine is working.

my working:

using F = ma (resolving parallel to slope):

2000gsina - 1600 = 0
2000gsina = 1600

i was planning to use P = Fv but this is not working. can anybody tell me why the equation i got above is wrong? im pretty sure i got the forces right

thanks
2. You can't say the resultant force is 0, if that were the case the engine would not be working.

P=fv

The force is 2000g/25 - 1600. Multiply that by 14 to get your answer in watts.
3. Just did the calculation doesn't seem right, is it moving up or down the slope?
4. (Original post by Ben121)
You can't say the resultant force is 0, if that were the case the engine would not be working.

P=fv

The force is 2000g/25 - 1600. Multiply that by 14 to get your answer in watts.
constant speed = 0 acceleration.

F = ma

when a = 0, so is the resultant force.
5. (Original post by Ben121)
Just did the calculation doesn't seem right, is it moving up or down the slope?
do i really need to answer this? lol
6. Oh bloody hell my head isn't in the game today, sorry ignore everything I've said.
7. (Original post by dildo baz00ka)
A lorry of mass 2000 kg is moving down a straight road inclined at angle a to the horizontal, where sina =1/25 . The resistance to motion is modelled as a constant force of magnitude 1600 N. The lorry is moving at a constant speed of 14 m s–1.

Find, in kW, the rate at which the lorry’s engine is working.

my working:

using F = ma (resolving parallel to slope):

2000gsina - 1600 = 0
2000gsina = 1600

i was planning to use P = Fv but this is not working. can anybody tell me why the equation i got above is wrong? im pretty sure i got the forces right

thanks
If it is moving up the slope, the resistive forces AND the component of its weight act against the object. For the lorry to be able to move up the slope the engine must be exerting a force, D. Hence, using F=ma,

D - 2000gsina - 1600 = 0

However, if it is moving down the slope, the the component adds to the tractive force of the engine and the resistive forces change direction. With this in mind, can you continue?
8. Ah hang on.

F + 2000gsina = 1600
F = 1600 - 80g

P = (1600-80g)*14 = 11424W = 11.424 kW
9. (Original post by Ben121)
Ah hang on.

F + 2000gsina = 1600
F = 1600 - 80g

P = (1600-80g)*14 = 11424W = 11.424 kW
I suggest you don't post full solutions.

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