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course PHY121
Week 5 Quiz 1 Version 2A ball starting from rest rolls 13 cm down an incline on which its acceleration is 25 cm/s2, then onto a second incline 30 cm long on which its acceleration is 7 cm/s2. How much time does it spend on each incline? For the first ramp:
vo = 0, a = 25cm/s^2, and `ds = 13cm
vf^2 = 0 + 2(25cm/s^2)(13cm)
vf^2 = sqrt 650cm^2/s^2
vf= sqrt 650cm^2/s^2
vf = 25.495 cm/s
`dt = `dv / a = (vf - vo)/a
`dt = (25.495cm/s - 0) / 25cm/s^2
`dt = 1.019 seconds
vf^2 = vo^2 + 2a`ds
vf^2 = (25.495cm/s)^2 + 2(7cm/s^2)(30cm)
vf^2 = 1070
vf = 32.711 cm/s
`dt = `dv / a = (vf - vo)/a
`dt = (32.711cm/s - 25.495cm/s) / 7cm/s^2
`dt = 1.031 seconds
Week 5 Quiz 2 Version 2
If the slope of a graph of the acceleration of a cart vs. the number of paper clips attached by a string and suspended over a pulley is ( 18 cm/s2) / clip, and if the slope of a graph of number of paper clips needed to maintain equilibrium vs. ramp slope is 59 clips / unit of ramp slope, then how many cm/s2 of acceleration should correspond to 1 unit of ramp slope? If 57 clips are necessary 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?18cm/s^2 / clip = S1
59 clips / unit of ramp slope = S2
_cm/s^2 = 1 unit RS
1 clip = 18cm/s^2
59 clips = 1 unit RS
1062 cm/s^2 = 1 unit RS
59*18 = 1062
57*18 = 1026
1062 - 1026 = 36 cm/s^2
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