# Motion due to gravity, mechanics help needed

• Nov 11th 2012, 07:59 AM
mysur
Motion due to gravity, mechanics help needed
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.
• Nov 11th 2012, 08:09 AM
skeeter
Re: Motion due to gravity, mechanics help needed
using Newton's 2nd Law, you should be able to determine the following magnitudes for acceleration.

no friction ...

$\displaystyle a = g\sin{\theta}$ directed down the track

with friction and the particle moving up the track ...

$\displaystyle a = g(\sin{\theta} + \mu \cos{\theta})$ directed down the track

with friction and the particle moving gown the track ...

$\displaystyle a = g(\sin{\theta} - \mu \cos{\theta})$ directed down the track

... the rest is all kinematics in one dimension.
• Nov 11th 2012, 08:12 AM
mysur
Re: Motion due to gravity, mechanics help needed
Thank you so very much
• Nov 11th 2012, 08:25 AM
mysur
Re: Motion due to gravity, mechanics help needed
But I get 9.8x sin-1(3/5) = 361.228N. Big numbers not sure
• Nov 11th 2012, 08:48 AM
skeeter
Re: Motion due to gravity, mechanics help needed
Quote:

Originally Posted by mysur
But I get 9.8x sin-1(3/5) = 361.228N. Big numbers not sure

note that $\displaystyle \theta = \sin^{-1}\left(\frac{3}{5}\right)$ , so $\displaystyle \sin{\theta} = \frac{3}{5}$

$\displaystyle a = g\sin{\theta} = \frac{3g}{5}$

kapish?
• Nov 11th 2012, 09:08 AM
mysur
Re: Motion due to gravity, mechanics help needed
oops
• Nov 11th 2012, 12:58 PM
HallsofIvy
Re: Motion due to gravity, mechanics help needed
That is, by the way, a "3- 4- 5 right triangle". The acceleration due to gravity, g= 9.8 m/s^2, is directed downward, parallel to the "3" side and so the component of acceleration down the hypotenuse, the "5" side, is -(3/5)g= -5.88 m/s^2. At that acceleration, the object will slow from 28 m/s to 0 in 28/5.88= 4.76 seconds. It will, in that time, move a distance -5.88(4.76)^2/2+ 28(4.76)= 66 and 2/3 meters up the ramp.
With no friction, it will return to the bottom in another 4.76 seconds, a total of 9.52 seconds from the instant it started moving and, of course, will have regained all its kinetic energy and so its speed, when it gets to the bottom of the ramp again, will be, again, 28 m/s.
• Nov 12th 2012, 12:05 PM
mysur
Re: Motion due to gravity, mechanics help needed
thank you Thank You Thank You