Answer:
D
Explanation:
From the information given:
The angular speed for the block [tex]\omega = 50 \ rad/s[/tex]
Disk radius (r) = 0.2 m
The block Initial velocity is:
[tex]v = r \omega \\ \\ v = (0.2 \times 50) \\ \\ v= 10 \ m/s[/tex]
Change in the block's angular speed is:
[tex]\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s[/tex]
However, on the disk, moment of inertIa is:
[tex]I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2[/tex]
The time t = 10s
∴
Frictional torques by the wall on the disk is:
[tex]T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10}) \\ \\ =0.6 \ N.m[/tex]
Finally, the frictional force is calculated as:
[tex]F = \dfrac{T}r{}[/tex]
[tex]F= \dfrac{0.6}{0.2} \\ \\ F = 3N[/tex]
