Answer: 0.0169J/s-m of force will be exerted per sec per meter displacement of the paper
Step-by-step explanation:
Power p = 3W
Distance r = 0.8m
Intensity = p/4¶r^2
Intensity = 3/(4x3.142x0.8^2) = 0.373W/m2
The paper is 8.5" by 11"
Converting to metre becomes
0.2032m by 0.2794m
Are of paper = 0.2032x0.2794 = 0.0568m2
80% of the light reflected is,
0.8x0.373 = 0.2984W/m2
Energy per sec In the reflected light is 0.2984x0.0568 = 0.0169J/s
Force = energy per sec per unit displacement of paper
= 0.0169J/s-m