Conservation of Energy

**craig** · March 28th 2010, 07:05 AM

Taking the value of $g$ to be $9.8 ms^{-2}$ .

Body of mass $5kg$ falls freely from rest a distance of $3m$ , find the kinetic energy acquired by this body.

Let the height above the ground be $h$ .

Initially $KE = 0$ , and Gravitational Energy (PE) = $mgh = 5gh$ .

Due to the conservation of energy, we know that this total of energy stays the same. So at our point 3 metres below the starting point:

$KE = 2.5v^2$ , $PE = 5g(h-3)$ .

Not sure how to work out the KE acquired by the body, and have I written the equations right?

Part 2: If the particle is brought to rest by a force of $70N$ , find the further distance fallen.

No idea where to start on this one, any help would be greatly appreciated.

Thanks again in advance.

**skeeter** · March 28th 2010, 07:50 AM

Body of mass 5kg falls freely from rest a distance of 3m, find the kinetic energy acquired by this body.

$\Delta KE = -\Delta GPE$

$\Delta KE = -(mgh_f - mgh_0) = -mg(\Delta h) = -mg(-3) = 3mg<br />$

go ahead and find the velocity ... you'll need it for part 2.

Part 2: If the particle is brought to rest by a force of 70N, find the further distance fallen.

$F_{net} = ma$

$70 - mg = ma$

$a = \frac{70 - 5g}{5}$

$v_f = 0$

now use the equation $v_f^2 = v_0^2 + 2a(\Delta y)$ to determine $\Delta y$ ... remember that $\Delta y < 0$

**craig** · March 28th 2010, 08:33 AM

Originally Posted by skeeter

Body of mass 5kg falls freely from rest a distance of 3m, find the kinetic energy acquired by this body.

$\Delta KE = -\Delta GPE$

$\Delta KE = -(mgh_f - mgh_0) = -mg(\Delta h) = -mg(-3) = 3mg<br />$

go ahead and find the velocity ... you'll need it for part 2.

Thanks for the reply.

For the velocity, $\frac{1}{2}mv^2 = 3mg.$ , so $v = \sqrt{6g} \approx 7.668$

Just to clarify, $h_f$ is the initial height?

Originally Posted by skeeter

Part 2: If the particle is brought to rest by a force of 70N, find the further distance fallen.

$F_{net} = ma$

$70 - mg = ma$

$a = \frac{70 - 5g}{5}$

$v_f = 0$

now use the equation $v_f^2 = v_0^2 + 2a(\Delta y)$ to determine $\Delta y$ ... remember that $\Delta y < 0$

So from your equation, $v_f^2 = v_0^2 + 2a(\Delta y)$ .

So $6g + 2(14 - g)(\Delta y) = 0$

$\Delta y = \frac{-6g}{2(14 - g)} = \frac{-3g}{14 - g} = -7$

So the further distance fallen is 7 metres?

**skeeter** · March 28th 2010, 08:42 AM

Originally Posted by craig

Thanks for the reply.

For the velocity, $\frac{1}{2}mv^2 = 3mg.$ , so $v = \sqrt{6g} \approx 7.668$

Just to clarify, $h_f$ is the initial height?

no, $\textcolor{red}{h_f}$ is final height

So from your equation, $v_f^2 = v_0^2 + 2a(\Delta y)$ .

So $6g + 2(14 - g)(\Delta y) = 0$

$\Delta y = \frac{-6g}{2(14 - g)} = \frac{-3g}{14 - g} = -7$

So the further distance fallen is 7 metres?

correct

...

**craig** · March 28th 2010, 08:44 AM

Thanks again.

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