tnhoots Posts: 28, Reputation: 1 New Member #1 Feb 18, 2007, 12:44 PM
Internal and Kinetic Energy.
1. The problem statement, all variables and given/known data
A 14.0 kg block is dragged over a rough, horizontal surface by a 78.0 N force acting at 20.0° above the horizontal. The block is displaced 5.00 m, and the coefficient of kinetic friction is 0.300.

(a) What is increase in internal energy of the block-surface system due to friction?

(b) Find the total change in the block's kinetic energy.

2. Relevant equations

(a) Fk=(coefficient of kinetic friction)(mass)(gravity)
Change in energy=(Fk)(displacement)

(b) Change in K= Work of other forces - Change in energy

3. The attempt at a solution

(a)Fk=(coefficient of kinetic friction)(mass)(gravity)=(.3)(14)(9.8)=41.16
Change in energy=(Fk)(displacement)=41.16(5)=205.8J

(b)Change in K= Work of other forces - Change in energy
= (78.0)(5)(cos20)
= 366.5 - 205.8=160.7J

however, these answers are wrong..? Suggestions?
 Nosnosna Posts: 434, Reputation: 103 Full Member #2 Feb 18, 2007, 12:57 PM
The force of gravity is only half of the story: Since the force being applied to the load isn't level, part of it is applied against gravity.

You'll need to decompose the force into its horizontal component and vertical component. You can do this by drawing a 20-70-90 right triangle with the hypotenuse length of 78.0. The shorter leg of the triangle will be your vertical component, and the longer will be the horizontal. Combine the vertical with gravity (taking care to remember that they are in opposite directions), and use the horizontal for the directional force.
 tnhoots Posts: 28, Reputation: 1 New Member #3 Feb 18, 2007, 01:50 PM
Wow, I'm just a little stumped with that explanation. Would you mind giving me an example?
 Nosnosna Posts: 434, Reputation: 103 Full Member #4 Feb 18, 2007, 01:59 PM
Can't give an example (coming up with examples is why I can never be a math instructor ;)), but I'll try to explain it better :)

The force of friction is based not on gravity, but by the force pushing the two objects together. In this case, you have gravity pulling the object down, and the applied force pulling it up.

How much it pulls it up is based on the angle from horizontal... in this case 20 degrees. Part of your 78N force goes towards pulling the object up, and part of it goes towards sliding the object horizontally. How much each of those parts gets will give you the forces needed in those equations.

Edit to add: Think about it like walking... if you head out northeast, and walk 1.4 miles, you can think of it as though you've walked 1 mile east and 1 mile north. Turning that into forces, your force of 1.4 northeast is made up of two component forces: 1.0 north and 1.0 east. Combining them gives you the overall applied force.
 tnhoots Posts: 28, Reputation: 1 New Member #5 Feb 18, 2007, 07:14 PM
Huh?
 Capuchin Posts: 5,255, Reputation: 656 Uber Member #6 Feb 19, 2007, 12:37 AM
The coefficient of friction is a value that resists horizontal force for every amount of vertical force.

If you are pushing at 90 degrees, then the only vertical force is the weight of the box. However if you are pushing at some other angle, you are either increasing or decreasing the vertical force, and so the box is either harder or easier to push.

You need to work out the total downwards force and put this into your equation.

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