Question

1)

Violet almost got caught the last time she took Victoria's box of red vines. But when Victoria gets a new 200 pound box of red vines (90.7 kg) for her birthday, Violet can't resist. As the party winds down and people begin to leave, Violet waits until Victoria goes to the bathroom before making her move. But with a box this big Violet decides to pull it instead. She ties the ribbons from Victoria's birthday presents together, wraps it around the box of red vines, and pulls with a force of 800 N at an angle of 30 degrees above the horizontal. If the coefficient of friction between the box and the carpet is 0.9, what is the acceleration of Victoria's box of red vines?

2)

Charge 1 is 200 μC and is at the origin (0 m, 0m)

Charge 2 is 75 μC and is at point (3 m, 4 m)

What is the force on Charge 2 in terms of its x (i) and y (j) components?

Answer 1

1) The force F by which violet pulls has to components one vertical F_y = Fsin30 and one horizontal F_x = Fcos30

Now; F_y = F/2 = 400N and F_x = F/2 = 400 N;

Also the force of gravity on it will be F_g = -90.7kg x 9.8m/s² = -888.86N

So, the normal force acting on the box will be

R = -(F_y + F_g) = 488.86N

The frictional force f = R = 0.9 x 488.86 N = 439.874N

So, the net force in horizontal becomes

F_net = 400 N - 439.974N

=> F_net = 252.846N

Thus, a_net = F_net/m = 2.79 m/s²

2) The position vector of 2 w.r.t 1 is given as r = (3,4) = 3 + 4 m

|r| = (3² + 4²)^1/2m = 5m

Now the force on charge 2 is given as

F = kQ₁Q₂r/|r|³

=> F = 9 x 10⁹ x 75C x 200C x (3 + 4 m)/5³

=> F = 1.08(3 + 4) N

=> F = 3.24 + 4.32 N