A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the disk is 5kg and its radius is 2 m. What is the linear speed, in m/s, after the mass drops by 0.3 m

1.98 m/s

Explanation:

To solve this, we would be using the law of conservation of energy, i.e total initial energy is equal to total final energy.

E(i) = E(f)

mgh = ½Iw² + ½mv²

Recall, v = wr, thus, w = v/r

Also, I = ½mr²

I = 0.5 * 5 * 2²

I = 10 kgm²

Remember,

mgh = ½Iw² + ½mv²

Substituting w for v/r, we have

mgh = ½I(v/r)² + ½mv²

Now, putting the values in the equation, we have

5 * 9.8 * 0.3 = ½ * 10 * (v/2)² + ½ * 5 * v²

14.7 = 1.25 v² + 2.5 v²

14.7 = 3.75 v²

v² = 14.7/3.75

v² = 3.92

v = √3.92

v = 1.98 m/s

Thus, the speed is 1.98 m/s

2b2t

explanation:

2b2t