$M = 70 \ kg$

$m = 10 \ kg$

$N = ?$

$r = 200 \ mm$

$F = 20 \ N$

$sin \ \alpha = \frac{200}{300} = 0.667$

$\alpha = 41.8°$

$sin \ \beta = \frac{200 – 40}{300}$

$\beta = 32.23°$

$q = \frac{tan \ \beta}{tan \ \alpha} = \frac{tan \ 32.23°}{tan \ 41.8°}$

$q = 0.705$

$h = \sqrt{300^2 – 200^2} = 0.2236 \ m$

w/o friction:

$0.2236 = \frac{895}{N^2} [ 1+ \frac{70}{2 \times 10} (1 + 0.705)]$

$N = 166.99 \ r.p.m.$

consider friction force of $20 \ N$

$h = \frac{895}{N^2} [1 + \frac{Mg + f}{2mg} (1 + q)]$

$0.2236 = \frac{895}{N^2} [1 + \frac{70 \times 9.81 – 20}{2 \times 10 \times 9.81} (1 + 0.705)]$

$N_1 = 164.9 \ r.p.m.$

$\frac{h}{0.2236} = \frac{895}{N^2} [ 1+ \frac{70 \times 9.81 + 20}{2 \times 10 \times 9.81} (1 + 0.705)]$

$N_2 = 169.06 \ r.p.m.$

Range, $N_2 – N_1 = 169.06 \ – 164.9$

$N_2 – N_1 = 4.16 \ r.p.m.$