Question

A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is fixed in place at the bottom of the incline. If the inclined plane makes an angle θ of 35.0 ∘ with the horizontal, what is the magnitude of the acceleration of the box when it is 61.0 cm from the bottom of the incline?

Answer 1

Answer:

$a=16.2m/s^{2}$

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

$F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}$

When fixed,F=o

Hence

$masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}$

The value of the acceleration is 16.2m/s^2