Find the compression in the spring with sprin | vedclass

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(A) The force exerted by radiation on a surface is given by $F = \frac{I A}{c} (1 + ho) \cos^2 	heta$,where $I$ is the intensity,$A$ is the area,$c

Vedclass · Accepted Answer

$0.5$

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(A) The force exerted by radiation on a surface is given by $F = \frac{I A}{c} (1 + ho) \cos^2 	heta$,where $I$ is the intensity,$A$ is the area,$c$ is the speed of light $(3 	imes 10^8 \ m/s)$,$ho$ is the reflection coefficient,and $	heta$ is the angle of incidence with the normal.Given: $I = 40 	imes 10^6 \ W/m^2$,$A = 50 \ m^2$,$ho = 0.5$,$	heta = 60^o$,$K = 5 \ N/m$.Calculating the force: $F = \frac{(40 	imes 10^6) 	imes 50}{3 	imes 10^8} 	imes (1 + 0.5) 	imes \cos^2(60^o)$.$F = \frac{2000 	imes 10^6}{3 	imes 10^8} 	imes 1.5 	imes (0.5)^2$.$F = \frac{20}{3} 	imes 1.5 	imes 0.25 = \frac{20}{3} 	imes 0.375 = 2.5 \ N$.Since $F = Kx$,the compression $x = \frac{F}{K} = \frac{2.5}{5} = 0.5 \ m$.

Find the compression in the spring with spring constant $K = 5 \ N/m$ which is attached to a metal plate of area $50 \ m^2$. Light of intensity $I = 40 \ MW/m^2$ is incident at an angle of $60^o$ on the metal plate,and the coefficient of reflection is $\rho = 0.5$ for the plate. (in $m$)

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