Thursday

November 27, 2014

November 27, 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.

**Answer this Question**

**Related Questions**

Physics - A skier of mass 79.1 kg, starting from rest, slides down a slope at an...

physics - A skier of mass 77.7 kg, starting from rest, slides down a slope at an...

eng physics - in a downhill race a skier slides down a 40 degree slope. starting...

physics - Starting from rest, a 75 kg skier slides down a 17.0° slope. If the ...

physics - A skier with a mass of 70 kg starts from rest and skis down an icy (...

Physics - A skier with a mass of 56 kg starts from rest and skis down an icy (...

Physics - A skier with a mass of 66.6 kg starts from rest and skis down an icy (...

PHYSICSSSSSS please help - An 60 kg skier is sliding down a ski slope at a ...

physics - A skier slides horizontally along the snow for a distance of 20.8 m ...

physics - A skier slides horizontally along the snow for a distance of 22.3 m ...