A0.23 kg ladle sliding on a horizontal frictionless surface is attached to one end of a horizontal spring (k = 580 n/m) whose other end is fixed. the ladle has a kinetic energy of 170 j as it passes through its equilibrium position (the point at which the spring force is zero). (a) at what rate is the spring doing work on the ladle as the ladle passes through its equilibrium position? (b) at what rate is the spring doing work on the ladle when the spring is compressed 0.66 m and the ladle is moving away from the equilibrium position?

Explanation:

Angular speed of the motion ( SHM )

ω = √k/m

= √(580/.23 )

= 50.20 radian /s

a ) Rate of doing work

= power = force x velocity

At the equilibrium position force becomes zero so

rate of doing work is zero.

b )

If a be the amplitude

1/2 k a² = 170

a = .7655 m

kinetic energy at equilibrium = 1/ 2 m v₀²

1/ 2 m v₀² = 170

.5 x  23  v₀² = 170

v₀ = 3.84 m /s which is the maximum velocity.

Given x = .66 where rate of doing work is to be calculated.

Force at x = ω² x

= 50.20² x .66 =

= 1663.22 N

Velocity v = v₀ √( a² - x² )

= 3.84 √( .7655² - .66 )

= 3.84 x .387

= 1.486 m/s

power = force x velocity

=  1663.22 x 1.486

= 2471.55  W .

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.