As $a=\frac{d v}{d t}=\frac{F}{m}=$ constant, hence, on integration, we get $;$ $\mathrm{v}=\frac{\mathrm{F}}{\mathrm{m}} \mathrm{t}$ So, power, $P

$F \propto \frac{1}{\upsilon }$

As $a=\frac{d v}{d t}=\frac{F}{m}=$ constant, hence, on integration, we get $;$ $\mathrm{v}=\frac{\mathrm{F}}{\mathrm{m}} \mathrm{t}$ So, power, $P=\mathrm{F} \mathrm{v}=\frac{\mathrm{F}^{2}}{\mathrm{m}} \mathrm{t}$ i.e., $p$ $\infty$ $t$ For power to be constant, $\mathbf{Fv} =$ constant, i.e., $\mathrm{F} \propto \frac{1}{\mathrm{v}}$

English

5.Work, Energy, Power and Collision

When a constant force is applied to a body moving with constant acceleration, power does not remain constant. For power to be constant, the force has to vary with speed as follows

A
$F \propto \frac{1}{\upsilon }$
B
$F \propto \frac{1}{\sqrt \upsilon }$
C
$F \propto {\upsilon }$
D
$F \propto {\upsilon ^2}$

Std 11
Physics

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When a constant force is applied to a body moving with constant acceleration, power does not remain constant. For power to be constant, the force has to vary with speed as follows

Similar Questions

The bob of a pendulum of length $l$ is pulled aside from its equilibrium position through an angle $\theta $ and then released. The bob will then pass through its equilibrium position with speed $v$ , where $v$ equals

A particle is moved from $(0, 0)$ to $(a, a)$ under a force $\vec F = (3\hat i + 4\hat j)$ from two paths. Path $1$ is $OP$ and path $2$ is $OQP$. Let $W_1$ and $W_2$ be the work done by this force in these two paths respectively. Then

A $15\, g$ ball is shot from a spring gun whose spring has a force constant of $600\, N\, m$. The spring is compressed by $3\, cm$. The greatest possible velocity of the ball for this compression is ............. $\mathrm{m}/ \mathrm{s}$ $(g = 10\, m/s^2$)

Two bodies of masses $m_1$ and $m_2$ are moving with same kinetic energy. If $P_1$ and $P_2$ are their respective momentum, the ratio $\frac{P_1}{P_2}$ is equal to