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olganol [36]
3 years ago
14

A 0.358 kg particle slides around a horizontal track. The track has a smooth, vertical outer wall forming a circle with a radius

of 1.66 m. The particle is given an initial speed of 8.19 m/s. After one revolution, its speed has dropped to 6.29 m/s because of friction with the rough floor of the track. Calculate the energy loss due to friction in one revolution.
Physics
1 answer:
oee [108]3 years ago
4 0

Answer:

The energy loss due to friction in one revolution is 4.92 J.

Explanation:

Given that,

Mass of particle = 0.358 kg

Radius = 1.66 m

Initial speed =8.19 m/s

Final speed = 6.29 m/s

We need to calculate the kinetic energy

Using formula of kinetic energy

\Delta K.E=\dfrac{1}{2}m(v_{f}^2-v_{i}^2)

Put the value into the formula

\Delta K.E=\dfrac{1}{2}\times0.358\times(8.19^2-6.29^2)

\Delta K.E=4.92\ J

Hence, The energy loss due to friction in one revolution is 4.92 J.

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t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

t=0.50s   (13) >>>This is the time it takes to the block to go from the initial velocity V_{o} to its final velocity V_{f}

If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

V_{f}-V_{o}=a.t   (13)

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V_{f}=(5.88m/{s}^{2})(0.50s)   (15)

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