Question

a small block on a frictionless, horizontal surface has a mass of 2.30×10^−2 kg. it is attached to a massless cord passing through a hole in the surface (figure 1). the block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.17 rad/s. the cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. model the block as a particle. what is the new angular speed? express your answer in radians per second. ω₂ = 8.68 rad/s previous answers correct part c find the change in kinetic energy of the block. express your answer with the appropriate units. δk = value units submit previous answers request answer x incorrect; try again; 5 attempts remaining you have found the final kinetic energy of the block. you need to determine the part d how much work was done in pulling the cord? express your answer with the appropriate units. w = value units units input for part d submit previous answers request answer

Explanation:

Step1: Calculate initial kinetic energy

The formula for kinetic energy of a rotating particle is \(K = \frac{1}{2}I\omega^{2}\), and for a particle \(I = mr^{2}\).
Initial kinetic energy \(K_{1}=\frac{1}{2}m r_{1}^{2}\omega_{1}^{2}\)
Given \(m = 2.30\times10^{- 2}\text{ kg}\), \(r_{1}=0.300\text{ m}\), \(\omega_{1}=2.17\text{ rad/s}\)
\(K_{1}=\frac{1}{2}\times(2.30\times 10^{-2})\times(0.300)^{2}\times(2.17)^{2}\)
\(K_{1}=\frac{1}{2}\times2.30\times10^{-2}\times0.09\times4.7089\)
\(K_{1}=4.89\times10^{-3}\text{ J}\)

Step2: Calculate final kinetic energy

Final kinetic energy \(K_{2}=\frac{1}{2}m r_{2}^{2}\omega_{2}^{2}\)
Given \(r_{2} = 0.150\text{ m}\), \(\omega_{2}=8.68\text{ rad/s}\)
\(K_{2}=\frac{1}{2}\times(2.30\times 10^{-2})\times(0.150)^{2}\times(8.68)^{2}\)
\(K_{2}=\frac{1}{2}\times2.30\times10^{-2}\times0.0225\times75.3424\)
\(K_{2}=1.96\times10^{-2}\text{ J}\)

Step3: Calculate change in kinetic energy

\(\Delta K=K_{2}-K_{1}\)
\(\Delta K = 1.96\times10^{-2}-4.89\times10^{-3}\)
\(\Delta K=1.47\times 10^{-2}\text{ J}\)

Step4: Use work - energy theorem for work done

According to work - energy theorem \(W=\Delta K\) (since the only force doing work is the tension in the cord and there is no non - conservative force other than tension, and the potential energy change is zero in this horizontal motion)
So \(W = 1.47\times10^{-2}\text{ J}\)

Answer:

For Part C: \(\Delta K = 1.47\times10^{-2}\text{ J}\)
For Part D: \(W=1.47\times10^{-2}\text{ J}\)