#$&*
course Phy 201
I was totally stumped by this question. Tried for 1.5hrs with no luck.
How many cm/s2 of acceleration should correspond to 1 unit of ramp slope if:•a graph of the acceleration of a cart vs. the number of paper clips attached by a string and suspended over a pulley has slope ( 19 cm/s2) / clip, and
• a graph of number of paper clips needed to maintain equilibrium vs. ramp slope has slope 53 clips / unit of ramp slope?
If 51 clips are required to match the mass of the cart, then if we could apply this force to the cart without the extra mass of all those clips, what would be the acceleration of the cart?
If for every clip the cart accelerates 19cm/sec2 and it takes 53clips to equal 1 unit of ramp slope then 19cm/sec^2 * 53(clips) = 1007cm/sec^2.
????????I drew out the first graph of 19cm/sec^2 for every 1 clip and then drew out a graph of the 53clips per unit of slope, but I could not understand how we would draw up the picture of the situation. Is it that we have a cart on a slope with the pulley at the end and the paperclip as weight on the other side???????
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Imagine that you have a cart, and a few paperclips in the cart. Suspend one of the clips over a low-mass, low-friction pulley. It will accelerate the cart, and you can take data to determine its acceleration. Add another clip to the first and the cart will accelerate more, this time with greater acceleration. Add another clip, and you'll get an even greater acceleration.
Graph acceleration vs. number of clips and fit a straight line to your graph. If the slope is 19 cm/s^2 / clip, and if it takes 53 clips to equal the mass of the system, then your conclusion is that a net force equal to the weight of the system will accelerate it at your 1007 cm/s^2.
This would then be your estimate of the acceleration of gravity.