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Vesna [10]
3 years ago
10

A 55 kg gymnast wedges himself between two closely spaced vertical walls by pressing his hands and feet against the walls. Part

a what is the magnitude of the friction force on each hand and foot? Assume they are all equal.
Physics
1 answer:
Aliun [14]3 years ago
5 0

solution:

Let frictional forces due to both hands and feets be "Ff" each(since its given that they all are equal), acting in upward direction( in opposite direction of supposed motion).\\
Then since there is no motion of gymnast thus net frictional force due to both hands and feets will be exactly balanced by the weight of the gymnast,\\  i.e\\
4f_{f}=weight =mg\\
f_{f}=\frac{55x9.8}{4}\\
=134.75N

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<em>where E_{ph}: is the photon energy, E_{0p} and E_{0ap}: are the rest energies of the proton and the antiproton, respectively, equals to m₀c², K_{p} and K_{ap}: are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.</em>        

Therefore the kinetic energy of the antiproton is:    

K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}

<u>The proton mass is equal to the antiproton mass, so</u>:

K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}  

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K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV

K_{ap} = 80.75 MeV              

Hence, the kinetic energy of the antiproton is 80.75 MeV.

I hope it helps you!

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