The initial velocity of a 1.90 kg block sliding down a frictionless inclined plane is found to be 1.10 m/s. Then 1.81 s later, it has a velocity of 7.13 m/s. What is the angle of the plane, with respect to the horizontal?
2007-10-02 06:36:08 · 3 answers · asked by jULIAN C 2 in Science & Mathematics ➔ Physics
Well, first find the block's acceleration. This is easy, because:
acceleration = (change in velocity) ⁄ (elapsed time)
Since they tell you the change in velocity and the elapsed time, just plug in the numbers.
Next, use the formula for (frictionless) acceleration down an inclined plane:
acceleration = g(sinθ)
You know the acceleration (from 1st formula); and you know "g". So solve for θ.
2007-10-02 06:41:59 · answer #1 · answered by RickB 7 · 0⤊ 0⤋
a million) internet rigidity = ma 60(cos20)-6.5(sin27) = 6.5(a) a=8.22 m/s/s (rounded down and would't locate the squared key for unit) i think of it particularly is nice yet i'm valuable i'm forgetting something... that is plenty much less complicated with a diagram however... Edit: certainly, I did. The 20 levels of the rope to be desirable. mounted.
2016-10-05 23:32:33 · answer #2 · answered by richberg 4 · 0⤊ 0⤋
along an incline plane acceleration a = g sin (angle)
Here, a = (v2-v1)/ t
= (7.13 - 1.10)/1.81
= 3.33 m/ss
So, 3.33 = 9.81 * sin (angle)
=> sin (angle) = 3.33/ 9.81 = 0.34
=> angle = 19.88 deg
I hope it helps
2007-10-02 07:50:14 · answer #3 · answered by Ehsan R 3 · 0⤊ 0⤋