Really? You Thought We had no HW?

I wrote this on the board before we started talking about the lab, but yes, we DO have a homework assignment for the weekend!

Chapter 6 - pages 172-179, Question 2 and 3, Problems 1,2,4,5,6, and 8.

Enjoy! Solutions to the HW and quiz will be posted later.

UPDATE:
Quiz solutions are posted here.

Solutions:

1. In this problem, assume the normal force is doing the work in equilibrium, so Fn - mg = 0, so Fn = mg. The work is therefore mg*delta_x* cos(1) = 7350 J
2.
a) Constant speed, so F - Fk = 0. The work done by F = (180 N)(6 m)(1) = 1080 J
b) Again constant speed, so F = mg. W = mg*delta_x*1 = 5400 J

4. The retarding (friction) force does negative work, preventing the car from speeding up. The work done by friction is negative since if the car moves to the right, friction is directed to the left. Thus the only unknown in the equation is the magnitude of the force. F = -70,000J/(2800m*(-1)) = 25 N

5. The only force acting on the rock is gravity. You know how much work is done on it, so the only unknown is the displacement. delta_x = W/mg = 36.1 meters. The signs of the work require a bit more explanation than will make sense here - we can talk more about it in class.

6. The maximum work is what would happen if all of the work done by gravity during the fall went into the nail. This means you have to find the work done by gravity as it falls. This is mg*delta_x*(1) = 7.84 J. This is not very much work! This is why it pays off to exert a bit more force when you are hammering, rather than let gravity do all the work.


7. Any time you are asked to find the minimum work, assume the work is done at equilibrium. Thus the force required to push the cart up the incline in (a) is mg sin(theta). The work is therefore mg*sin(17.5)*300m*(1) = 884,000 J

b) Again, using sigma_Fx = 0, the force required to do this is (mg*sin(17.5) + mu*mg*cos(17.5))* 300 m = 1,600,000 J.

8. Normal force and gravity do no work since they are perpendicular to the horizontal displacement. Since the cart travels at constant speed, the sum of the forces in both directions is zero.

F cos(20) - Fk = 0 so Fk = F cos(20).

The work done by F is F cos(20)* 15 m *(+1) = 169.1 J.

The work done by Fk = F cos(20)* 15 m * cos(180) = -169.1 J.