Finding Mass and Speed in Kinetic Energy Problems

In summary, the conversation discusses two challenging problems - one involving the mass and velocity of two runners and the other involving a truck traveling up and down a hill. Despite attempting to equate kinetic energies, the correct solution for the first problem was not achieved. The second problem proves to be the toughest of the semester and the conversation seeks guidance on how to approach it.
  • #1
destro47
16
0
Having trouble with the following two part problem:

A) Joe, mass 89.7 kg, is racing against Tom. When Joe and Tom have the same kinetic energy, Tom is running faster. When Joe increases his speed by 22.8%, they are running at the same speed. Find Tom's mass.

Tried equating their kinetic energies and solving for the unknown mass, but that did not work. The reasoning behind it seemed logical, I was surprised that the answer turned out to be wrong.

B) A highway goes up a hill, rising at a constant rate of 1.00 m for every 50 m along the road. A truck climbs this hill at constant speed vup = 16 m/s, against a resisting force (friction) equal to 1/25 of the weight of the truck. Now the truck comes down the hill, using the same power as it did going up. Find vdown, the constant speed with which the truck comes down the hill.
ASSUME: the resisting force (friction) has the same magnitude going up as going down.

This one is really tough, probably the hardest all semester. All I know is the slope of the highway going up, 1/50, and that the coefficient of friction is a fraction of the force needed to move the car at constant v.

I would appreciate any guidance on this one, to steer my thoughts in the right direction. I am not just looking for solutions here.
 
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  • #2
A) Can you show your work for A? I think you're on the right track with setting their kinetic energy equal, (and then using what you know about Joe's velocity in terms of Tom's velocity) but I'm not sure how your method didn't work.
 
  • #3
1/2 mv^2 = 1/2 mv^2
1/2 m (.228v)^2 =1/2 mv^2
1/2 m .052v^2 = 1/2 mv^2
.052 (89.7) = m
 
  • #4
destro47. You'll need to distinguish their masses and velocities:
e.g. 1/2 M1 V1^2 = 1/2 M2 V2^2...
 
  • #5
the left side of the equation is Joe and the right is Tom
 
  • #6
Since Joe has to increase his velocity by 22.8% to match speed with Tom, you want 1.228vJoe = vTom, not .228vJoe=vTom.
 
  • #7
destro47 said:
the left side of the equation is Joe and the right is Tom

Ok but using sides of the equation to distinguish variables is very very very bad notation. You can't then apply any of the usual algebraic techniques. If you want to avoid subscripts let one of two's variables be represented with capital letters.
 

Related to Finding Mass and Speed in Kinetic Energy Problems

1. What is work?

Work is defined as the transfer of energy that results from a force applied over a certain distance. It is typically measured in joules (J) and is calculated by multiplying the force applied by the distance over which the force acts.

2. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is dependent on the mass and velocity of the object and is calculated by the equation KE = 1/2 * m * v^2, where m is the mass and v is the velocity.

3. How are work and kinetic energy related?

Work and kinetic energy are directly related. This means that the work done on an object is equal to the change in kinetic energy of the object. In other words, when work is done on an object, its kinetic energy changes by the same amount.

4. Can work and kinetic energy be negative?

Yes, both work and kinetic energy can be negative. Negative work occurs when the force and displacement are in opposite directions, while negative kinetic energy can occur when an object is slowing down or moving in the opposite direction of its initial velocity.

5. What are some real-world examples of work and kinetic energy?

Some examples of work and kinetic energy in everyday life include pushing a shopping cart, riding a bike, and throwing a ball. In each of these cases, work is done on the object, and its kinetic energy changes as a result. Other examples include moving objects with a crane, running, and jumping on a trampoline.

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