A ball starts from rest and accelerates at 0.520 m/s 2 while moving down an incl

Question

A ball starts from rest and accelerates at 0.520 m/s2 while moving down an inclined plane 8.45 m long. When it reaches the bottom, the ball rolls up another plane, where, after moving 14.65 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 8.30 m along the second plane?

Explanation / Answer

Given,

Initial velocity, u = 0

A)

Using relation, v^2 = u^2 + 2as

Velocity, v = sqrt (2as) = sqrt (2 x 0.52 x 8.45) = 2.96 m/s

B)

Time, t = (v-u) /a = 2.96/0.52 = 5.7 sec

C)

Acceleration, a = v^2/2s = - (2.96^2)/(2 x 14.65) = - 0.30 m/s^2

Magnitude is 0.30 m/s^2, negative sign shows deceleration

D)

Velocity, v^2 = (2.96)^2 + 2(-0.30)(8.30) = 1.94 m/s

Comment in case any doubt please rate my answer ....

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.