Answer:
The work done on the object by the force in the 5.60 s interval is 40.93 J.
Explanation:
Given that,
Force
Mass of object = 2.00 kg
Initial position
Final position
Time = 4.00 sec
We need to calculate the work done on the object by the force in the 5.60 s interval.
Using formula of work done
Put the value into the formula
Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.
Answer:
Revolutions made before attaining angular velocity of 30 rad/s:
θ = 3.92 revolutions
Explanation:
Given that:
L(final) = 10.7 kgm²/s
L(initial) = 0
time = 8s
<h3>Find Torque:</h3>Torque is the rate of change of angular momentum:
We know that
T = Iα
α = T/I
where I = moment of inertia = 2.2kgm²
α = 1.34/2.2
α = 0.61 rad/s²
<h3>Find Time 't'</h3>
We know that angular equation of motion is:
ω²(final) = ω²(initial) +2αθ
(30 rad/s)² = 0 + 2(0.61 rad/s²)θ
θ = (30 rad/s)²/ 2(0.61 rad/s²)
θ = 24.6 radians
Convert it into revolutions:
θ = 24.6/ 2π
θ = 3.92 revolutions