Working out speed, kinetic energy and resistive force

AI Thread Summary
The discussion focuses on calculating speed, kinetic energy, and resistive forces in a physics problem. Participants emphasize the importance of understanding gravitational potential energy and kinetic energy formulas, particularly kinetic energy = 1/2 × mass × speed². They explore the relationship between work, force, and distance, noting that work done equals the frictional force multiplied by the distance traveled. Clarifications are made regarding the constancy of force in magnitude, even if it varies as a vector. The conversation concludes with confirmation that all necessary information is available to determine the required magnitudes.
Meezus
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Homework Statement


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Homework Equations


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The Attempt at a Solution



a) I'm not sure how to calculate speed without having both distance and time but I believe it will revolve around calculating gravitational potential energy and kinetic energy.

bi) I think this is just simply using the formula kinetic energy = 1⁄2 × mass × speed2.

bii/biii) I'm really struggling to work out these. I think It might revolve around using the speed from the first question and then take it away from 22m/s.
 
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You are on the right track for a) and b.i).
I suggest you solve those two first. It might then be more apparent how to solve b.ii)
 
haruspex said:
You are on the right track for a) and b.i).
I suggest you solve those two first. It might then be more apparent how to solve b.ii)

I believe I have solved them now.
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if possible could you help with the last one?
 
Meezus said:
if possible could you help with the last one?
What equation do you know relating work and distance?
 
haruspex said:
What equation do you know relating work and distance?

Work = Force x Distance.?
 
Meezus said:
Work = Force x Distance.?
Looks promising. How would it apply here? Explain in words.
 
haruspex said:
Looks promising. How would it apply here? Explain in words.

is it something like:
Work done = change in kinetic energy,
Work = - 11250 J?
 
Meezus said:
is it something like:
Work done = change in kinetic energy,
Work = - 11250 J?
Yes, but I meant the work = force x distance equation. Can you express that in respect of frictional force and the circumstances in this question?
 
haruspex said:
Yes, but I meant the work = force x distance equation. Can you express that in respect of frictional force and the circumstances in this question?

The amount of work done is equal to the frictional force times by the distance? I'm sorry I'm not 100% sure.
 
  • #10
Meezus said:
The amount of work done is equal to the frictional force times by the distance? I'm sorry I'm not 100% sure.
Yes (given that the force is constant; in general the relationship is an integral).
 
  • #11
haruspex said:
Yes (given that the force is constant; in general the relationship is an integral).

Am I able to work out the magnitude from this?
 
  • #12
Meezus said:
Am I able to work out the magnitude from this?
Yes, you have all the information. (One clarification: the force is not constant as a vector here, but it is constant in magnitude. This works out ok because the force of friction is always parallel to the motion, so it still reduces to force x distance traveled along the path. You do not know and do not need to know the end-to-end displacement as a vector.)
 
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