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tigry1 [53]
4 years ago
8

A spool of thin wire (with inner radius r = 0.45 m, outer radius R = 0.60 m, and moment of inertia Icm = 0.8208 kg·m2) pivots on

a shaft. The wire is pulled down by a mass M = 1.300 kg. After falling a distance D = 0.500 m, starting from rest, the mass has a speed of v = 63.1 cm/s. Calculate the energy lost to friction during that time.
Physics
1 answer:
makkiz [27]4 years ago
3 0

Answer:

E_L=4.667J

Explanation:

The energy lost can be model by the energy equation in potential and kinetic energy in each step of the motion

Given: r=0.45m,R=0.60m,I_m=0.8208 kg*m^2,v=63.1cm/s, m=1.3kg

E_p=m*g*r

E_p=1.3kg*9.8m/s^2*0.45m

E_p=5.733 J

Kinetic energy:

K_{Es}=\frac{1}{2}*I*w^2, K_{Ep}=\frac{1}{2}*m*v^2

w=v^2/r

Energy lost :

E_p-K_{Es}-K_{Ep}=0

5.733J-\frac{1}{2}*0.8208kg*(\frac{(0.631m/s)^2}{0.45m})^2-\frac{1}{2}*1.3kg*(0.631m/s)^2=0

E_L=4.667J

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The wavelength of the golf ball is <u>2.328×10⁻³⁴m.</u>

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The deBroglie wavelength λ of a particle is given by,

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Calculate the deBroglie wavelength of the moving golf ball by substituting 6.626×10⁻³⁴J s for h, 45.9×10⁻³kg for m and 62.0 m/s for v.

\lambda=\frac{h}{mv}\\ =\frac{6.626*10^-^3^4Js}{(45.9*10^-^3kg)(62.0m/s)} \\ =2.328*10^-^3^4m

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4 years ago
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