Tuesday

July 22, 2014

July 22, 2014

Posted by **James** on Monday, May 27, 2013 at 12:29pm.

- Physics -
**MathMate**, Monday, May 27, 2013 at 6:40pmCase 1: zero friction

F=mg sin(28)

a=F/m=g sin(28)

S=ut+(1/2)at²

100=0+g sin(28)t²/2

t=sqrt(2*100/(g sin(28))

=6.59 s.

Case 2: μ_{k}=0.17

Frictional force

=μmg cos(28)

Net force (assuming μcos<sin)

=mg(sin(28)-μcos(28)

t=sqrt(2*100/(g (sin(28)-μcos(28))

=7.99 s

Take the difference.

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