A block is given an initial velocity of 2.00 m/s up a frictionless 22.0° incline. How far up the incline does the block slide before coming (momentarily) to rest?
Im sure this is most likely a very simple problem, however I just dont even know where to start! Thanks in advance!
2007-10-10 08:36:56 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
V final =0
0 = 2.0 -9.8sin22*t
t= 0.6 seconds
distance up the incline = d = v 0*t -9sin22*t^2/2
d= 2*t - (9.8*sin22*t^2)/2 = 0.53 meters
2007-10-10 09:16:40 · answer #1 · answered by lonelyspirit 5 · 0⤊ 0⤋
If you have worked on the idea of conservation of energy, you can do this that way.
Initial KE = (1/2)m*v^2
After it comes to a stop, all that Initial KE has been converted to potential energy.
U = m*g*h
Set the 2 energy expressions equal to each other and solve for h. The h will be measured along a vertical line. You'll need to do some trig to find out how high up the incline that is.
If you haven't gotten that far, you need to find the component of the acceleration due to gravity that points down the slope -- opposite the original velocity. Then use
Vf^2 = Vo^2 + 2*a*s
Where Vf = 0 when it comes to rest, you were given Vo , you calculated a, and s is the upslope distance it goes.
2007-10-10 16:39:45 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
Draw a free body diagram. Divide the acceleration of gravity into x and y components with x acting parallel to the incline and y perpendicular to the incline. Assuming no friction, gravity is the only force deccelerating the block.
2007-10-10 15:43:09 · answer #3 · answered by Fireball 3 · 1⤊ 0⤋