Sunday

April 20, 2014

April 20, 2014

Posted by **Ed** on Sunday, March 10, 2013 at 11:46am.

I'm not looking for a solution which includes Potential Energy. We haven't learned that in class yet, is there a another way to solve this?

- Physics -
**Damon**, Sunday, March 10, 2013 at 1:19pmThe energy way is the easy way but.

Normal force = m g cos 14

force down ramp = m g sin 14 + mu m g cos 14 = m g (mu cos 14+ sin 14)

= m g ( 1.12 )

F = m a

a = F/m = g (1.12) = 10.9 m/s^2 down ramp

say t is time to stop

v = Vi - 10.9 t

0 = Vi - 10.9 t

so

t = Vi/10.9

d = Vi t - (10.9/2) t^2

85 = Vi^2/10.9 - (10.9/2)(Vi/10.9)^2

85 = (Vi^2/10.9 )( 1 - 1/2)

170 = Vi^2/10.9

Vi = 43 m/s

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