Answer:
D
Step-by-step explanation:
From the information given:
The angular speed for the block
![\omega = 50 \ rad/s](https://img.qammunity.org/2022/formulas/physics/college/1gzqe1xd6wmx9aj9s4j08w23sqqnw12wqf.png)
Disk radius (r) = 0.2 m
The block Initial velocity is:
![v = r \omega \\ \\ v = (0.2 * 50) \\ \\ v= 10 \ m/s](https://img.qammunity.org/2022/formulas/physics/college/17inlpgv8umz4a8mc18acgjryomtkfggkm.png)
Change in the block's angular speed is:
![\Delta _(\omega) = \omega - 0 \\ \\ = 50 \ rad/s](https://img.qammunity.org/2022/formulas/physics/college/rann962e3jcv2n3xukihq6r8erebfuaikq.png)
However, on the disk, moment of inertIa is:
![I= mr^2 \\ \\ I = (3 * 0.2^2) \\ \\ I = 0.12 \ kgm^2](https://img.qammunity.org/2022/formulas/physics/college/9mqjjrtymb0ytc3bvvzwr3rguywa7jwvj1.png)
The time t = 10s
∴
Frictional torques by the wall on the disk is:
![T = I * ((\Delta_(\omega))/(t)) \\ \\ = 0.12 * ((50)/(10)) \\ \\ =0.6 \ N.m](https://img.qammunity.org/2022/formulas/physics/college/itwx0wpjqpxwgbtbb2bwe6pauxgrfwkcjp.png)
Finally, the frictional force is calculated as:
![F = \frac{T}r{}](https://img.qammunity.org/2022/formulas/physics/college/6qurx19l1k8lqm6ii85vasagmlf6h97v4t.png)
![F= (0.6)/(0.2) \\ \\ F = 3N](https://img.qammunity.org/2022/formulas/physics/college/8n5qcozm9c2oa8abwagl2wgpv7id7y6n8c.png)