hello478
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- Homework Statement
- question c part ii) power produced by engine
- Relevant Equations
- p= f* v
The discussion revolves around the interaction of power, force, and velocity in the context of a lorry moving up a slope. Participants are examining how to calculate the power produced by the engine, considering the forces acting on the vehicle, including resistive forces and gravitational components.
The discussion is ongoing, with participants exploring various interpretations of the forces involved and how they relate to the power produced by the engine. Some guidance has been offered regarding the relationship between force and velocity, but no consensus has been reached on the specific calculations or assumptions.
Participants note the importance of understanding the problem context, including the mass of the lorry and the slope, while also expressing concerns about missing information and the need for clarity in the problem statement.
its by finding the component of weight along slopeBvU said:8628.46 ?
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i have attached the question image above..., is it not there? let me check againBvU said:So the slope is given, and so is the mass of the lorry. Given for you , that is.
But my telepathic capabilities are limited and I need a complete problem statement, typed out.
And: what are the directions of 8628.46 and the resistive force ?
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The F you need to use in P=Fv is the motive force delivered by the engine. This has to balance the opposing forces, the 5200N of resistance plus the gravitational component.hello478 said:Homework Statement: question c part ii) power produced by engine
Relevant Equations: p= f* v
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ok so in p=fvharuspex said:The F you need to use in P=Fv is the motive force delivered by the engine. This has to balance the opposing forces, the 5200N of resistance plus the gravitational component.
Depends what you mean by "total force". As you wrote, since the speed is constant the total force on the truck is zero.hello478 said:ok so in p=fv
the f would always be total force?
yeah the net force would be zeroharuspex said:Depends what you mean by "total force". As you wrote, since the speed is constant the total force on the truck is zero.
Yes, that being the static frictional force on the tyres. We can consider that the motive force provided by the engine, though in reality what the engine provides is a torque.hello478 said:yeah the net force would be zero
but here... we would have to use the force which is causing the body to move
Not sure what you mean by that. In the general case the truck is accelerating, so we have ##F_{net}=F_{static}+F_{resist}=ma##, ##F_{resist}## being negative. So the engine power would be ##F_{static}v=(ma-F_{resist})v##.hello478 said:(the total force, sum of resistive forces) with that velocity???
is this explanation fine?
i meant that we would multiply the force which causes to move the lorry with the velocity it createsharuspex said:Not sure what you mean by that.
You don't have to care what creates the velocity. What matters is that a force ##\vec F## is exerted at a point on a body while that point is moving with velocity ##\vec v##. The power transferred is ##\vec F\cdot\vec v##.hello478 said:i meant that we would multiply the force which causes to move the lorry with the velocity it creates