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Topic review (newest first)
- bob bundy
- 2012-11-12 10:54:51
Hi Mosur,
Welcome to the forum.
I like to start with a drawing with the forces shown in red. (see below)
I've got the force due to gravity as mg
the normal reaction of the slope as R
and I've put in a frictional force F even though it isn't needed for the first part.
So with F = 0
the acceleration down the slope is a = mg times 3/5
you know the initial velocity u = 28
it will stop going up when its velocity v = 0
so you can get the distance, s, using
and for the time, t
On the way back u will now be zero, 'a' is reversed, s is unchanged so v is easy to get.
When there's a frictional force you'll need to resolve along the slope and perpendicular to it
That's for the travel up the slope.
On the way back the friction will oppose the motion, so will be up the slope.
Bob
- Mosur
- 2012-11-12 06:36:44
A smooth track is inclined to the horizontal at an angle of sin-1(3/5). A particle of mass m is placed on the track at a point O, and projected directly up the track with speed 28ms-1. Find how far the particle goes up the track, the speed at which it returnsto O, and the time that it takes from the instant of projection until it gets back to O. What would the corresponding answers be if the plane were rough with coefficient of friction 9/20?
thank you very much in advance.