Question

A pulley having a 0.20-m radius is turning at 200 rad/s around a central axis. It slowed down uniformly to 125 rad/s in 4-seconds. 1. A. Compute the angular acceleration of the pulley B. Compute the angle through which the pulley turns in this time. C. Compute the length of belt it winds in that same time 2. How much work is done on a 3.0-kg block of ice in moving it from rest to a speed of 4.0 m/s?

Answer 1

angular acceleration = -18.75 rad/s²

angle = 650 rad

length = 130 m

work done = 24 J

Step-by-step explanation:

given data

radius = 0.20 m

speed s1 = 200 rad/s

speed s2 = 125 rad/s

time = 4 s

mass = 3 kg

speed = 4 m/s

to find out

angular acceleration, angle and length and work done

solution

we apply angular acceleration formula that is

s2 = s1 + αt

put here value

125 = 200 + α(4)

α = -18.75 rad/s²

angular acceleration = -18.75 rad/s²

and

angle will calculated as

angle = s1(t) + 1/2 × α × t²

angle = 200(4) + 1/2 × (-18.75) × 4²

angle = 650 rad

and

length will be

length = radius × angle

length = 0.2 × 650

length = 130 m

and

work done is calculated as

work done = 1/2 × mv²

work done = 1/2 × 3(4)²

work done = 24 J