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2 years ago

8

After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of vi=2.62 m/s

. To reach the rack, the ball rolls up a ramp that rises through a vertical distance of h=0.47m. What is the linear speed of the ball when it reaches the top of the ramp?

1 answer:

Misha Larkins [42]2 years ago

5 0

Answer:

vf = 4.01 m/s

Explanation:

According to the law of conservation of energy:

$Kinetic\ Energy\ Lost\ by\ Ball = Potential\ Energy\ Gained\ by\ the\ Ball \\\frac{1}{2}m(v_f^2-v_i^2) = mgh\\\\ v_f^2-v_i^2 = 2gh\\v_f^2 = 2gh + v_i^2\\$

where,

vf = final speed of ball at top of the ramp = ?

vi = initial speed of ball = 2.62 m/s

g = acceleration due to gravity = 9.81 m/s²

h = height = 0.47 m

Therefore,

$v_f^2 = (2)(9.81\ m/s^2)(0.47\ m)+(2.62\ m/s)^2\\v_f = \sqrt{9.2214\ m^2/s^2+6.8644\ m^2/s^2}\\$

<u>vf = 4.01 m/s</u>

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The force of friction is 300 N

Explanation:

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where

$\sum F$ is the net force acting on the object

m is its mass

a is its acceleration

For the cart in this problem, we have two forces acting on it:

- The force of push, F = 500 N, forward

- The force of friction, $F_f$ , backward

So Newton's second law can be rewritten as

$F-F_f = ma$

where

m = 50 kg

$a=4 m/s^2$ is the acceleration of the cart

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Answer:

The girl has greater tangential acceleration

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Since all the points of the merry go round complete 1 circle in the same time, the angular velocity of each point of the merry go round is the same, and so all the points also have the same angular acceleration.

The tangential acceleration instead is given by

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Since the girl is near the outer edge and the boy is closer to the centre, the value of r for the girl is larger than for the boy, so the girl has greater tangential acceleration.

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Answer:

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Answer:

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<u>Instant Acceleration</u>

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