Powerful forces

Classical Mechanics Level 3

A constant power $P$ is applied to a particle of mass $m$ . Neglecting friction, when the particle's velocity changes from $u$ to $v$ , what is the distance travelled by the particle?

\frac{m}{P}(v^{2}-u^{2})

\frac{m}{P}(v-u)

\frac{m}{3P}(v^{2}-u^{2})

\frac{m}{3P}(v^{3}-u^{3})

5 solutions

Agnishom Chattopadhyay
May 19, 2014

I do not know how to do the problem but I figured it would be the last one by dimensional analysis.

Power,P=F.v=mav=mv(v.dv/ds)

$ds=(mv^{2}dv)/P$ .Integrate to get the answer.

Snehdeep Arora - 7 years ago

$P = F v \underbrace{=}_{F = m a} m a v \underbrace{=}_{a = \frac{dv}{dt}} m v \frac{dv}{dt} = m v \frac{dv}{ds}\frac{ds}{dt} \underbrace{=}_{\frac{ds}{dt} = v} m v^2 \frac{dv}{ds}$

$\Rightarrow ds = \frac{m}{P} v^2 dv$

$\Rightarrow \int_0^{s_{final}} ds = \int_u^v \frac{m}{P} v^2 dv$

$\Rightarrow s_{final} = \frac{m}{3P} (v^3 - u^3)$

Émile Staël - 7 years ago

thats awesome.

Ishan Mahmud - 6 years, 10 months ago

The units are fine. If you work from the relation $\ P \ = \ \vec{F} \cdot \vec{v}$ , the quantities in the expression are

$\frac{M}{Mav} \ \cdot \ v^3 \ = \ \frac{v^2}{a} \ = \ \frac{m^2 / sec^2}{m/sec^2} \ = \ m$ .

Working with fundamental units, we have

$\frac{kg}{W} \ \cdot \ \frac{m^3}{sec^3} \ = \ \frac{kg}{J/sec} \ \cdot \ \frac{m^3}{sec^3} \ = \ \frac{1}{J} \ \cdot \ \frac{kg \ m^2}{sec^2} \ \cdot \ m \ = \ \frac{1}{J} \ \cdot \ J \ \cdot \ m \ = \ m \ \$ .

Gregory Ruffa - 7 years ago

The units don't match here

Simona Vesela - 7 years ago

same here i did....

Saurav Sharma - 7 years ago

i also did the same thing

Murlidhar Sharma - 7 years ago

still i m getting confused sir.....

Rehanullah Khan - 7 years ago

P = F v = m a v; also, a = dv/dt and v = dx/dt => dx = [m (v^2) dv]/P so, integral [dx] = x = (m/3P) (v^3 - u^3)

Shin Shi - 7 years ago

not completly understand

H Hassan Mustafa - 7 years ago

Same with me . I tried only dimensional analysis.

Niranjan Khanderia - 6 years, 11 months ago

the equation is dimensionally wrong.........mass divided by power alone gives the dimensions of acceleration times velocity.....hw is this possible

Sayam Chakravarty - 7 years ago

There is nothing wrong: $a \times v = \frac {m}{s^{2}} \times \frac {m}{s} = \frac{m^{2}}{s^{3}}$

The last option is: $\frac {kg \times v^{3}}{ \frac {3 kg \times m^{2}}{s^{3}} } = \frac {\frac {m^{3}}{s^{3}}}{\frac {m^{2}}{s^{3}}} = m$

When you divide the cube of the velocity by kg/W you end up with meters.

Petru Lupsac - 7 years ago

yes it is dimensionally wrong

Namit Yadav - 7 years ago

Let s be the distance covered by the particle, P be the force to the mass m. So P=mf f = accelaration. we know f= v^2-u^2/2s... now putting it to P=mf we get tthe ans, m/2P(v^2-u^20

Rikan Banerjee - 7 years ago

Gregory Ruffa
May 19, 2014

The reason there won't be a very simple approach to obtaining the distance is that the acceleration of the particle isn't constant: applying a constant power to the particle means that it is kinetic energy that will be changing by equal amounts in equal intervals of time. So the velocity of the particle cannot be following a linear function.

If we take the motion of the particle to be along a straight line (the argument can be modified for circular or more general paths), the applied power is $P \ = \ Fv \ = \ (ma) \ v \ = \ mv \ \frac{dv}{dt}$ . This gives us a separable differential equation for the velocity function $\ v(t) \$ . If we take $\ v(0) \ = \ U \$ and choose T to represent the time at which $\ v(T) \ = \ V \$ , we have

$v(t) dv \ = \ \frac{P}{m} \ dt \ \ \Rightarrow \ \ \int_U^v \ v(t) \ dv \ \ = \ \ \int_0^T \ \frac{P}{m} \ dt \ \ = \ \ \frac{1}{2} v^2 - \frac{1}{2} U^2 \ = \ \frac{P}{m} T \ \$ .

This gives us the length of time to change from speed U to speed V as $\ T \ = \ \frac{m}{2P} (V^2 - U^2) \$ . We also find that the velocity as a function of time is given by

$[ \ v(t) \ ]^2 \ = \ \frac{2P}{m} t \ + \ U^2 \ \ \Rightarrow \ \ v(t) \ = \ [ \ \frac{2P}{m} t \ + \ U^2 \ ]^{1/2} \ \$ .

This is what we shall integrate over time to find the distance the particle moves during the specified speed change:

$x \ = \ \int_0^T \ v(t) \ dt \ \ = \ \ \int_0^T \ [ \ \frac{2P}{m} t \ + \ U^2 \ ]^{1/2} \ dt \ \ = \ \left( \frac{2}{3} [ \ \frac{2P}{m} t \ + \ U^2 \ ]^{3/2} \cdot \frac{m}{2P} \right) \vert^T_0$

$= \ \frac{m}{3P}\left( [ \ \frac{2P}{m} T \ + \ U^2 \ ]^{3/2} \ - \ [ \ \frac{2P}{m} \cdot 0 \ + \ U^2 \ ]^{3/2}\right) \ = \ \frac{m}{3P}\left( ( \ \frac{2P}{m} \cdot \frac{m}{2P} \ [V^2 - U^2] \ + \ U^2 \ ) \ ^{3/2} \ - \ ( \ U^2 \ )^{3/2} \right)$

$= \ \frac{m}{3P}\left( [ \ V^2 \ ]^{3/2} \ - \ [ \ U^2 \ ]^{3/2} \right) \ = \ \frac{m}{3P}( \ V^3 \ - \ U^3 \ ) \ \$ .

Mass x accl x velocity = P Hence mv dv/dt =P (As power equals dot product of velocity and force and f=Ma.Note we have assumed body travels in a straight line). writing dv/dt as dv/ds multiplied by ds/dt equivalent to dv/ds multiplied by velocity (chain rule ), and integrating with limits we get the desired result.

Sridhar Thiagarajan - 7 years ago

I have a question. We use the formula P=Fv in case of constant motion but in this case the velocity of the matter changes from u to v. So what is the validity of using this formula?

Tanjim Faruk - 7 years ago

Power is a rate of energy transfer, so it is an instantaneous rate of change and is not limited to "constant motion". The argument I gave uses one-dimensional ("straight-line") motion with the force F acting along that line simply to reduce the "dot product" in the power expression to $\ P \ = \ \vec{F} \cdot \vec{v} \ = \ Fv \$ , but the argument can be generalized for other sorts of paths. In this expression, v is not necessarily a constant, so P can vary with time; in this problem, P is taken to be constant. The use of "v" in the answer is unfortunate, since we are also using it to represent the velocity function for the particle. Because of this, I have revised my solution to use U and V as the initial and final speed, keeping "v" for the function.

Gregory Ruffa - 7 years ago

Come on man !! Did you reallly have to find velocity as a function of time first? When we have P = mav just write a as dv/dx * dx/dt . Done!

Nikhil Mohan - 7 years ago

Brilliant!

Arsalan Iqbal - 7 years ago

Pratham Pandey
May 20, 2014

As we know that $P = F.v$ and we also know that $a = v{ dv }/{ dx }$ so we get $P=m{ v }^{ 2 }\ { dv }/{ dx }$ on further evaluation we get $\int _{ 0 }^{ s }{ dx } = \frac { m }{ p } \int _{ u }^{ v }{ { v }^{ 2 }dv } \\ \Rightarrow s = \frac { m }{ 3P } ({ v }^{ 3 } - { u }^{ 3 })$

Siddhant Pradhan
May 20, 2014

P=F.v P=(ma).v P/m=av we know, a=v.dv/dx P/m = v^2dv/dx dx= m/P v^2dv Integrate with limits to get answer

Arijit Banerjee
May 20, 2014

Use Dimensional Analysis ... u will get the answer

Powerful forces

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