(D) Given:Velocity $v = 7.2 \text{ km/hr} = 7.2 \times \frac{5}{18} = 2 \text{ m/s}$.Slope is $1$ in $20$, so $\sin \theta = \frac{1}{20}$.Mass $m = 1

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(D) Given:Velocity $v = 7.2 \text{ km/hr} = 7.2 \times \frac{5}{18} = 2 \text{ m/s}$.Slope is $1$ in $20$, so $\sin \theta = \frac{1}{20}$.Mass $m = 100 \text{ kg}$.Acceleration due to gravity $g = 9.8 \text{ m/s}^2$.When the man and cycle move up the hill, the component of weight opposing the motion is $F = mg \sin \theta$.The power $P$ required is given by $P = F \times v$.$P = (mg \sin \theta) \times v$.$P = 100 \times 9.8 \times \left( \frac{1}{20} \right) \times 2$.$P = 98 \text{ W}$.

Class 11 Physics Work, Energy and Power Power

Medium

$A$ man is riding a cycle with a velocity of $7.2 \text{ km/hr}$ up a hill having a slope of $1$ in $20$. The total mass of the man and the cycle is $100 \text{ kg}$. The power of the man is ............ $\text{W}$. (in $\text{ W}$)

A
$200$
B
$175$
C
$125$
D
$98$

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Medium

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$A$ man is riding a cycle with a velocity of $7.2 \text{ km/hr}$ up a hill having a slope of $1$ in $20$. The total mass of the man and the cycle is $100 \text{ kg}$. The power of the man is ............ $\text{W}$. (in $\text{ W}$)

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$A$ force applied by an engine on a train of mass $2.05 \times 10^6 \ kg$ changes its velocity from $5 \ m/s$ to $25 \ m/s$ in $5 \ minutes$. The power of the engine is (in $MW$)

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