Steven+Gambino

1. i. 10.59 m 1. ii. 8.33 2. i. 3.19 s 2. ii. 12.1 m/s Professor's Note : very Good effort - Give more explanation !! and try to avoid numerical values in the final answer -- check my page -- February 6, 2012 i) A = 5i + -3j ; B = i + 4j ii) |A| = 5.83 ; |B| = 4.12 iii) 106.94 degrees


 * My A and B are reversed**

Professor's Note : Job well done - still don't want to see numericals Week 5


 * a) Start by using the equation y-y1=Viyt+.5at^2**
 * Solve for t**
 * Plug t into change in position over change in time for the x direction to get the velocity of the pelican.**
 * b) Solve for time again, but replace 5.4m with 2.7. Plug the new time, which should be less than the first, into the same equation as part a and solve for the displacement in the x direction.**


 * Is this all we are supposed to do? Are we not supposed to post our solutions and work? If you go through my page, you will see what I am expecting. I am **__trying (by giving the best shot)**__ to increase the critical thinking abilities here ! I like you to explain how a problem is done, and somewhere else you can use the numbers and do the work and get the

Week 7

a) The balance will still read 65 N when the scale is moving at a constant velocity because the acceleration is still equal to zero.

b) Start by solving for the mass by dividing 65/9.8. The force on the balance is the difference between the net force and the force of gravity and can be found by multiplying the mass by the difference between the acceleration due to gravity and the net acceleration of the system.

Professor's Note: Good job !, You can use equations to do the explanations (e.g., W = mg --> m = W/g, Fnet = ma etc.)

Week 8

a) Because we need to prevent the block from sliding, it is a good idea to start by finding the force of static friction. Ff=uN. We can find the normal force by finding the component of the weight of the block. 85cos20 Multiply that result with u to get the force of friction. 85cos(20)*.25 Now we know the the there must be at least one other force acting against the block, gravity. So we can solve for the force of gravity by 85sin20. When solving for this value we can see that it is greater than the force of friction, meaning the force of friction alone will not keep the block up, resisting gravity. So the block needs a "push" up the incline, a third force. Because we have to add this third force the force of friction is going to oppose it as well as gravity, instead of acting with it. So using Fnet=0 (because there is no motion), we can see that the the third required force is gravity + friction. So : 85cos(20)*.25+85sin(20). b) In order to begin moving the block up the incline, all we need to do is add an infinitely small amount of force to oppose gravity and friction from problem (a). This will cause Fnet>0 and give the block a net acceleration up the incline. c) Part (c) is very similar to problem a because Fnet=0 because of the constant velocity. The only difference is the block is now moving so the friction used would be the kinetic friction not the static friction. 85cos(20)*.15+85sin(20)

Week 9

b)Wgravity=mgd a) Wnet=m*(g/4)*d Wcord=Wgravity-Wnet c) dKE=Wnet. Since it starts from rest the initial velocity, and therefore the initial KE, are equal to zero. So KE=Wnet. d) Using 1/2mv^2 and solving for v from the answer to part (c), you can find that v=sqrt(2Wnet/mass)


 * Week 10**

Because the collision is elastic we know two things. 1) momentum is conserved. 2) kinetic energy is conserved. Therefore we are able to form two equations: 1) (m1)(v1i)+(m2)(v2i)=(m1)(v1f)+(m2)(v2f) 2) (.5)(m1)(v1i)^2+(.5)(m2)(v2i)^2=(.5)(m1)(v1f)^2+(.5)(m2)(v2f)^2

At this point we also have three unknowns: the initial and final velocities of the ball and the final velocity of the block. We can narrow this down to two based on our understanding of centripetal force. In this case, at the bottom of the ball's arc, the only two forces acting on the object are mg and and the tension, which is also the normal force. Because the ball is not accelerating in the vertical direction we know the Fnet=0. Therefore, mg+T=0. T acts in the opposite direction as mg so will be considered negative. You can set these two forces equal to each other giving you: T=mg. Because T is also the centripetal force we can say: (mv^2)/r=mg. The mass cancel out and we solve for the blocks initial velocity immediately prior to the collision: v=sqrt(gr) Now when we plug that in to the two equations, we have to unknowns and two equations, a solvable system. Solve the system to give you the two unknown velocities.


 * GOOD JOB Steve, so far you are the only one who got the right answer !! way to go .... professor **