Posted by Charlene on Friday, March 23, 2007 at 3:00am.
picture in your mind a block that slides without friction down a ramp that is inclined at an angle (q) above the horizontal. Initially the block is at rest a distance (s) in meters (measured along the plane) from the bottom. Calculate the time (in seconds) required for the block to slide to the bottom.
can some1 show me the formula to find the time? for some reason, i can't find it in my textbook.
For Further Reading
* physics - drwls, Friday, March 23, 2007 at 8:04am
With no friction, the weight component down the ramp is Mg sin A, where g is the angle. That equals M a, so the acceleration is a = g sin A
The time T required is given by
(1/2)a T^2 = (1/2)g sin A
Solve for T
ookk. but what if there is no weight or mass of the block given? and only the angle and length? which is: 1 degrees & 300 meters. i have to find T in seconds.. would i do this?
(1/2)a T^2 = (1/2)g sin A
(1/2)(-.999)(T^2) = (1/2)(1)sin(300)
?? im still confused. help!
Please note the mass divides out, you don't need it.
a= gSinA
s= 1/2 a t^2 put in a, then solve for t.
t= sqrt 2s/a = sqrt (2s/gSinA)
thanks!
.. i kept gettin' the wrong answers.. and i was doing everything correctly!! then i realized my calculator was set to radians instead of degrees. arggh! lol
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