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Romashka-Z-Leto [24]
3 years ago
14

In 1994, Leroy Burrell of the United States set what was then a new world record for the men’s 100 m run. He ran the 1.00  102

m distance in 9.5 s. Assuming that he ran with a constant speed equal to his average speed, and his kinetic energy was 3.40  103 J, what was Burrell’s mass?
Physics
1 answer:
vovikov84 [41]3 years ago
5 0

Answer:

61.33 Kg

Explanation:

From the question given above, the following data were obtained:

Distance = 1×10² m

Time = 9.5 s

Kinetic energy (KE) = 3.40×10³ J

Mass (m) =?

Next, we shall determine the velocity Leroy Burrell. This can be obtained as follow:

Distance = 1×10² m

Time = 9.5 s

Velocity =?

Velocity = Distance / time

Velocity = 1×10² / 9.5

Velocity = 10.53 m/s

Finally, we shall determine the mass of Leroy Burrell. This can be obtained as follow:

Kinetic energy (KE) = 3.40×10³ J

Velocity (v) = 10.53 m/s

Mass (m) =?

KE = ½mv²

3.40×10³ = ½ × m × 10.53²

3.40×10³ = ½ × m × 110.8809

3.40×10³ = m × 55.44045

Divide both side by 55.44045

m = 3.40×10³ / 55.44045

m = 61.33 Kg

Thus, the mass of Leroy Burrell is 61.33 Kg

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\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since \Delta x=0, so:

E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2

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m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m

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E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J

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I hope this helps!
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Frequency of vibration for the arm = \frac{1}{2\pi}\sqrt\frac{9.8}{0.3933}

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Frequency of vibration for the arm = 0.7945 Hz

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