The work done \( W_f \) by the force \( F = 20 \, N \) is given by: \[ W_f = F \cdot d = 20 \times 1 = 20 \, J \] This is the change in kinetic energy of the wheel: \[ KE = \frac{1}{2} I \omega^2 \] Using \( I = MR^2 \) where \( M = 10 \, \text{kg} \) and \( R = 0.1 \, \text{m} \): \[ I = 10 \times (0.1)^2 = 0.1 \, \text{kg m}^2 \] Now equating the work done to the change in kinetic energy: \[ 20 = \frac{1}{2} \times 0.1 \times \omega^2 \quad \Rightarrow \quad \omega = 20 \, \text{rad/s} \]
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: