Aworker drags a crate across a factory floor by pulling on a rope tied to the crate (fig. 6-60). the worker exerts a force of 450 n on the rope, which is inclined at 38� to the horizontal, and the floor exerts a horizontal force of 125 n that opposes the motion.

(a) calculate the magnitude of the acceleration of the crate if its mass is 325 kg.
m/s2
(b) calculate the acceleration if the crate's weight is 325 n.
m/s2

(a) 0.71 m/s^2

We are only interested in the horizontal motion of the crate, so there are two forces acting on the crate along this direction:

- The horizontal component of the force exerted by the worker through the rope, which is given by

where F = 450 N and

- The horizontal force of friction that opposes the motion, given by

The two forces have opposite directions, so we must take into account a negative sign. According to Newton's second law, the resultant of these forces is equal to the product between the mass of the crate, m, and its acceleration, a:

where m = 325 kg. Re-arranging the equation, we can find the magnitude of the acceleration:

(b) 6.92 m/s^2

In this case, we need to find the mass of the crate first. We know that its weight is given by

where g=9.8 m/s^2 is the acceleration due to gravity. Since we know W=325 N, we find

And now we can use the same equation used in part (a) to find the acceleration of the crate:

small = small current large mag large current

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j