A ball starts from rest and accelerates at 0.530 m/s 2 while moving down an incl

Question

A ball starts from rest and accelerates at 0.530 m/s2 while moving down an inclined plane 9.90 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 14.90 m, it comes to rest. (Assume positive direction points down the first plane and up the second plane.)

(a) What is the speed of the ball at the bottom of the first plane?

(b) How long does it take to roll down the first plane?

(c) What is the acceleration along the second plane? (Careful with sign!)

(d) What is the ball's speed 7.50 m along the second plane?

(e) Draw position, velocity, and acceleration graphs.

Explanation / Answer

a)

here by using the third equation of motion

v^2 - u^2 = 2 *a *s

v = sqrt( 2 * 0.53 * 9.9)

v = 3.24 m/s

b)

use the first equation of motion

v = u + a*t

3.24 = 0 + 0.53 * t

t = 6.11 sec

c)

use the third equation of motion

v^2 - u^2 = 2 *a *s

0 - 3.24^2 = 2 * a * 14.9

a = - 0.35 m/s^2

d)

use the third equation of motion

v^2 - u^2 = 2 *a *s

v^2 = 3.24^2 + 2 * -0.35 * 7.5

v = 2.29 m/s

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