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3 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]3 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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4 years ago

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3 years ago

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Where our new initial position is $y_{max}$ , the velocity when the engine breaks is $v_e=v_i+at=120\frac{m}{s}$ and the only acceleration comes from gravity (which points down).

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3 years ago

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Answer:

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For the rock dropped by Bill, as the only influence on it is gravity (accelerating it downwards with an acceleration equal to g), and v₀ =0, we can use the following kinematic equation:

$y = \frac{1}{2} * g * t^{2}$

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As all the parameters are given in SI units, it is advisable to convert this value to m, as follows:

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Now, we can solve for t, as follows:

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