Posted by Sandhya on Thursday, March 27, 2008 at 4:03am.
I will be happy to critique your thinking.
Perhaps if you divide up the questions to different postings and add your answers, we will then be able to comment.
1. What is the best way to determine levelness?
See if slope of Velocity versus Time for both directions are equal but opposite.
2. In chronological order, what happens to the kinetic, potential, and total energy of the cart for one half cycle. The half cycle starts just after you have pushed the cart. The half cycle finishes just when the cart stops at the height of its motion. Remember, this is about the cart's mechancial energy.
Kinetic goes from maximum to zero (at the highest point of the incline.)
Potential goes from zero to a maximum slightly less than Kinetic's maximum.
Total Energy stays the same, but decreases slightly due to friction
3. What word (four letters long that starts with a w) describes the transfer of potential energy into kinetic energy?
4. Which situation has the greatest net force along incline?
when the cart is at rest
5. What are the units for energy?
All three units described here are valid for Energy
6. Let's say you measure an average acceleration to be 0.4241 ms-2.
What is the height (in centimeters) of the riser block?
Assume the legs of the track span 1.0000 m and that there is neither friction nor drag. Also assume the block is under one of the legs.
Answer not known
7. You have a level track. You push a cart with mass = 0.88[kg].
You measure the initial velocity to be 0.78[m s-1].
2 seconds later, you measure the velocity to be 0.585[m s-1].
What is the work (reported in mJ) that friction did on the cart?
work =200 [mJ]
8. You have a different system of unknown cart mass upon a level surface. The cart travels 55 [cm] in an unknown time period. The change in Kinetic Energy is -0.117117 [J]. What is the force of friction measured in Newtons?
N/A sufficient data not given
6 I dont understand
7 I didnt get that, it looks as if you rounded to one sig digit
f*.55=change in KE.
a = v-u / t = (0.585 - 0.780)/2 = 0.0975ms-2
S = ut + at2/2 = 0.78*2 - (0.0975*4)/2 = 1.56 - 0.195 = 1.36Work done by friction = m*a*S = 0.88*0.0975*1.365 =0.117117 J = 117.117 mJ
Will anyone help in solving prob.6?
for problem six you just divide the average acceleration multiplied by 1.0m by gravity.
(accelgrav)(1.0m)/(9.8) = You should get 4.12 or something :)
Question 4 is:
When the cart is going uphill. In this scenario (for this lab) the force that pushed the cart uphill is no longer in play. So the only forces acting on the cart are mgSin(theta) and friction, both of which are pointing in the same direction when the cart is going up hill, giving it the highest overall net force.
Question 1The best way to determine levelness is to see if slope of velocity versus time for both directions are equal but opposite.
Question 4:The greatest magnitude of net force is when the cart is going uphill. Uphill has a Net force greater by twice friction. It is not when the cart is at rest.
Question 6: mgsin(theta) = ma. Use to solve for theta, the angle. Then take the tangent of the angle to solve for h.
Question 8: There is enough info. Take the change in kinetic energy given in the problem and divide by the distance in the problem using the correct units. If the answer is in Newtons change your distance given in cm to the distance in meters before dividing.
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