Answer:
Expression of work done is

Work done to move the sled is given as 187.2 J
Explanation:
As we know that the formula of work done is given as

here we know that
F = 12.6 N
d = 15.4 m

so we will have


Answer:
Magnitude the net torque about its axis of rotation is 2.41 Nm
Solution:
As per the question:
The radius of the wrapped rope around the drum, r = 1.33 m
Force applied to the right side of the drum, F = 4.35 N
The radius of the rope wrapped around the core, r' = 0.51 m
Force on the cylinder in the downward direction, F' = 6.62 N
Now, the magnitude of the net torque is given by:

where
= Torque due to Force, F
= Torque due to Force, F'


Now,


The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.
Now, the magnitude of the net torque:

Answer:
a) Θ = ω₀*t + ½αt² To complete first revolution 2π rads = 0*t + ½αt² and to complete the first and second combined 4π rads = 0*t + ½α(t+0.810s)² Divide second by first: 2 = (t + 0.810s)² / t² This is quadratic in t and has roots at t = -0.336 s ← ignore and t = 1.96 s ◄ b) Use either equation from above: 2π rads = 0*t + ½α(1.96s)² α = 3.27 rad/s² ◄ Hope this helps!
Explanation: