Final questions

Good questions. See my notes and let me know if you have further questions.

course Phy 201

These are just some of the questions that keep popping up that I'm not 100% sure if I'm doing right or not. I have a feeling I need to do rather well on this final to get a C in the class so I'm sorry for bothering you with all these questions, but I'm never sure if I've done everything right... Thanks.

You did well on these questions.
Let me know if anything is unclear, and include specifics about what you do and do not understand.

1) Coasting from rest down a certain hill, whose slope is variable, I reach a speed of 17.49m/s at the bottom. If I coast from rest down the second half of the hill I reach a speed of 9m/s. Ignoring the effects of air resistance and friction: How fast would I therefore be going if I coasted from rest down the first half of th How high would I have to climb from the halfway point to reach the top? ***** I found in the book where these problems are worked out using the conservation of energy equation: 1/2mv1^2 + mgy1 = 1/2mv2^2 + mgy2 but in the book, the height of the hill is always given. How do you solve for velocity without knowing the height or distance of the hill?*****

You can find the KE for going down the entire hill, and you can find the KE for going down the second half. The difference is the KE you gain going down the first half.
This difference is equal to the PE loss down the first half.

2) The load in the back of my pickup truck will remain stationary if I slow from 22mph to rest in 4.8 seconds, but not if I slow any more quickly. What is the coefficient of static friction between the load and the truckbed? *****I couldn't find much in the book but tables with the coefficients of friction listed, but online I found this definition and equation: The static friction coefficient (μ) between two solid surfaces is defined as the ratio of the tangential force (F) required to produce sliding divided by the normal force between the surfaces (N) μ = F /N For a horizontal surface the horizontal force (F) to move a solid resting on a flat surface is F= μ x mass of solid x g. I wanted to make sure that this was a legit explanation and that the equations were correct. The thing is, even if they are the right equations, I'm not sure how to work the problem...*****

Find the acceleration required to slow from 22 mph to rest in 4.8 sec..
Find the net force required to provide this acceleration to the box.
The frictional force is the only force acting in the direction of motion, so it must exert this much force.
The normal force and the weight of the box will be equal and opposite in this case. So you can find the normal force.
frictional force = coeff of friction * normal force. So if you have the rest, you can easily find the coeff of friction.

3) A simple harmonic oscillator with mass 1.46kg and restoring force constant 430N/m is released from rest at a displacement of .55m from its equilibrium position. What will be its position at the clock time t = .3366s? What is the force on the oscillator at this position? *****This question also asked for the acceleration, which I tried by finding omega = sqrt(430N/m / 1.45kg) = 17.16rad/s * .55m = 9.44m/s.

omega is the angular frequency of the motion, and omega * r is the speed of the point on the reference circle. However, it is equal to the speed of the object only when the point on the reference circle is moving parallel to the oscillator.
If you use this to find the centripetal acceleration of the point on the reference circle, you will correctly obtain 162 m/s^2. However this is the acceleration of the oscillator only when it is at one of its extreme positions, when its distance from equilibrium is equal to the amplitude of motion.

Accel = v^2 / r = (9.44m/s)^2 / .55m = 162m/s^2. I wanted to make sure I'd gotten this much correct and then to see how to get the position at that clock time.

I thought of using the equation theta = omega0 'dt + .5 a *dt^2, which gives .5(162m/s^2)(.3366s)^2 = 9.12m, but I wanted to make sure that was the right way.

The equation you quote would apply to a uniform acceleration situation, but SHM does not involve uniform acceleration. So the equations of uniformly accelerated motion never apply here.

As for the friction at that specific point, I'm not sure because using simply F = m*a doesn't seem to consider one specific point...******

You have found omega. The motion is characterized by the equation
y = A cos(omega * t) = .55 m cos(17.16 rad/s * t).
If you plug in t = .3366 s you will get the position y.
Then using your force constant you can find the restoring force
F = - k y.

4) A simple harmonic oscillator of mass 40kg has a period of .04seconds. If the amplitude of its motion is 16560m, what are the maximum magnitudes of its acceleration and velocity? At what displacements from equilibrium can each maximum occur? *****I'm not sure if I'm using the right equations for the maximum magnitudes for velocity and acceleration. All I could find were vmax = omega A sin (omega 'dt) and accel = omega^2 A cos(omega 'dt). Using them, I got extremely large numbers (284,398rad/s and 4.057*10^8rad/s^2) and am not sure if this is correct or not.

vMax = omega * A and aMax = omega^2 * A, and your calculations look right. Something that moves through an oscillation with amplitude 16560 m in .04 seconds is really moving.

Also, as mentioned in the previous question, I'm not sure how to find the displacement at each maximum...*****

Max velocity occurs when the oscillator passes thru equilibrium, at displacement 0.
Max acceleration occurs at the extreme positions, where displacement are +16560 m and - 16560 m.

5) How do you find the KE and PE at equilibrium and halfway to equilibrium? I found a problem like this in the class notes, but I could not get any of the work I did in my calculator to equal the answers that the notes show so I'm not sure if I'm doing anything right or not. Here's a question that I'm having problems with: If a simple harmonic oscillator of mass .61kg is subjected to a restoring force of 3.9N when displaced .0427m from equilibrium, what will be its KE and its PE at equilibrium and halfway to equilibrium if it is released from rest at a displacement of .3904m from equilibrium? ***** I tried using PE = Fnet * 'ds = 3.9N * .9427m = .167J

This would not apply because the force would be changing.

and I also tried PE = kx^2, k = (.61kg * 9.8m/s^2) / .1952m = 30.63N/m

This is the correct expression for the PE. However F = -k x, so k = -F / x = -(-3.9 N) / .0427 m.

= 30.63N/m * (.1952m)^2 = 1.17J, so wouldn't that mean that KE = -1.17J? *****

The PE at max displacement is 1/2 k A^2. As the oscillator moves from position A to position x its PE decreases from 1/2 k A^2 to 1/2 k x^2.
The KE at max displacement is 0, since the object comes to rest for an instant at that position. The KE gain between position A and position x is equal to the PE loss between those positions.
So you can find the KE at position x.
Setting this equal to 1/2 m v^2 you get the velocity at position x.