The flywheel of a motor has a mass of 300.0 kg and a radius of 55 cm. If the motor can exert a torque of 2000.0 N . m on this flywheel, how much work does the motor do on the flywheel in the first 23 seconds?

Explanation:

Work done by torque is given as

Word one = torque × angular displacement

W = τ × θ

Given that,

τ = 2000Nm

Mass of motor = 300kg

Radius r = 55cm = 0.55m

Work done by wheel in first t= 23second.

Now we need to find the angular displacement

We know that,

τ = I•α

Moment of inertia of wheel

I = MR²

I = 300 × 0.55²

I = 90.75 kgm²

Then, τ = I•α

α = τ / I

α = 2000/90.75

α = 22.04rad/s²

Then, using circular motion,

∆θ = wit + ½αt²

wi = 0rad/s

∆θ = ½αt²

∆θ = ½ × 22.04 × 23²

∆θ = 5829.2 rad.

Then,

Work done?

W = τ × θ

W = 2000 × 5829.2

W = 1.17 × 10 ^7 J

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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answer(20) psi;