course PHY 231
7/15 2234
A ball starting from rest rolls 11 cm down an incline on which its acceleration is 26 cm/s2, then onto a second incline 39 cm long on which its acceleration is constant, and on which it spends 1.36 seconds. How much time does it spend on the first incline and what is its acceleration on the second?For the first incline:
`ds = v0 `dt + .5 a `dt^2
'ds = 11cm
a = 26cm/s^2
v0 = 0m/s
11cm = 0m*'dt/s + 13cm*'dt^2/s^2
11s^2/13 = 'dt^2
'dt = 0.92s (positive since this happens after the ball was released)
'ds/'dt = v_avg = 11.96cm/s ~= 12cm/s
2v_Avg = vf = 24cm/s
vfa = v0b = 24cm/s
'ds = 39cm
'dt = 1.36s
v_avg = 28.7cm/s = 29cm/s
vf = 2v_avg - v0 = 2*29cm/s - 24cm/s = 34cm/s
'dv = 10cm/s
a = 'dv/'dt = 7.35cm/s^2
The ball spends 0.92s on the first incline and has an acceleration of 7.35cm/s^2 on the second incline.
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