A ball starts from rest and accelerates at 0.500 m/s2 while moving down an inclined plane 9.00 m long. when it reaches the bottom, the ball rolls up another plane, where it comes to rest after moving 15.0m on that plane. (a) what is the speed of the ball at the bottom of the first plane? (b) during what time interval does the ball roll down the first plane? (c) what is the acceleration along the second plane? (d) what is the ball’s speed 8.00 m along the second plane?

a) final speed = ((initial speed) ^2+2(accelerarion)(distance)) ^0.5
= ((0)^2+2(0.05)(9))^0.5=0.949m/s

b) distance=1/2(initial velocity+final velocity) (time)
time = distance/0.5(initial velocity+final velocity) = 9/0.5(0.949)=19.0s

c)( (0.949)^2-0)/30=0.03m/s^2

d)( (0.949)^2-speed^2)/2*8=0.03
speed= 0.649m/s

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aristocles aristocles · Answer 2 · 2019-07-23T03:49:49+02:00

Answer:

Part a)

v = 3 m/s

Part b)

t = 6 s

Part c)

a = - 0.3 m/s/s

Part d)

v = 2.05 m/s

Explanation:

Part a)

as we know that initially ball starts from rest and move along the inclined plane with acceleration a = 0.5 m/s/s for distance d = 9 m

so here final speed is given as

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 0 = 2(0.500)(9)[/tex]

[tex]v_f = 3 m/s[/tex]

Part b)

For the time to slide down we will have

[tex]v_f - v_i = at[/tex]

[tex]3 - 0 = 0.5 t[/tex]

t = 6 s

part c)

for the other plane it stops after moving 15 m distance

so here we have

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]0 - 3^2 = 2(a)(15)[/tex]

[tex]a = -0.3 m/s^2[/tex]

Part d)

final speed of the ball after moving up by d = 8 m

[tex]v_f^2 - v_i^2 = 2 a d[/tex]

[tex]v_f^2 - 3^2 = 2(-0.3)(8)[/tex]

[tex]v_f = 2.05 m/s[/tex]