# Normal forces, Work, and Energy help!

1. Feb 22, 2008

### katye

a cart loaded with bricks has a total mass of 17.2 kg and is pulled at constant speed by a rope. the rope is inclined at 20.2 degrees above the horizontal and the cart moves 19.2 m on a horizontal floor. the coeffiecient of kinetic friction between the ground and the cart is .8. The acceleration of gravity is 9.8 m/s^2.

what is the normal force exerted on the cart by the floor?
how much work is done on the cart by the rope?
(note the energy change due to friction is a loss of energy) what is the energy change due to friction?

=basically i don;t even know where to start :(

any help is appreciated!
Thanks

2. Feb 22, 2008

### hage567

Start by drawing a free body diagram with all of the forces on it. Resolve the applied force into its components. Sum up the forces in each direction.

3. Feb 22, 2008

### katye

i've done that but i can't figure out how to solve for the applied force's x and y components because i can't set the sum of the forces in the x direction to zero because the box is moving right?

4. Feb 22, 2008

### hage567

Yes you can because it's not accelerating. It's moving at a constant speed so the net force must be zero.

5. Feb 22, 2008

### katye

alright that make since.
but my homework service keeps telling me i';m wrong. do you see any mistakes

F(x)=0=Xcos(20.2) - (.8*17.2*9.8)
and i get X=143.68567
then i use that for
F(y)=0=N+(17.2)(-9.8)+ (143.68567sin(20.2))
and i get N=118.9456

6. Feb 22, 2008

### hage567

This isn't quite right. The frictional force is $$f = \mu_kN$$, but you don't know what N is yet! That's what you are trying to solve for. In this case N is NOT simply equal to mg. There is a vertical component to the applied force so that affects the normal force. You must leave N as a variable in the first equation.

Your second equation is more or less right except for the 143.68567 part. You don't know what the force is yet so you must leave it as F (or X or whatever you are calling it).

So you have two equations and two unknowns (the applied force, and N), so you can solve for N with what you have.

7. Feb 22, 2008

### katye

oh my gosh. thank you so much. i know it seems really simple, but that problem has been killing me. i really appreciate it!

8. Feb 23, 2008

### katye

sorry, I'm trying to solve for how much work the rope is doing now and i thought the equation would just be
frictional force=applied force so (frictional coeff)*(Normal force)*(distance) but i was wrong, wrong, wrong. so .8(130.229)(19.2)
because 130.229 was what i found for my first answer for normal force

9. Feb 23, 2008

### katye

nevermind i got it! thanks!

10. Feb 23, 2008

### katye

and just for anyone who tries to use this in the future, the answer is the normal force you found in the first question*cos(angle)*distance :) best of luck!