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blsea [12.9K]
3 years ago
7

4. A latch holding a cart in place, hanging high above the ground, breaks and the cart falls. The cart started from rest and is

in free fall until it hits the ground.
a. What is the velocity after 3.2 s?



b. How far has the cart fallen at 3.2 s?
Physics
1 answer:
scoray [572]3 years ago
7 0

Answer:

a) 31.4 m/s

b) 50.2 m

Explanation:

a) When an object is free falling, its speed is determined by the gravity force giving it acceleration. Equation for the velocity of free fall started from the rest is:

v = g • t

g - is gravitational acceleration which is 9.81 m/s^2, sometimes rounded to 10

t - is the time of free fall

So:

v = 9.81 m/s^2 • 3.2

v = 31.4 m/s ( if g is rounded to 10, then the velocity is 10 • 3.2 = 32 m/s)

b) To determine the distance crossed in free fall we use the equation:

s = v0 + gt^2/2

v0 - is the starting velocity (since object started fall from rest, its v0 is 0)

s = gt^2/2

s = 9.81 m/s^2 • 3.2^2 / 2

s = 50.2 m (if we round g to 10 then the distance is 10 • 3.2^2/2 = 51.2 meters)

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m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

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The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

5 0
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