Kinetic Energy - College Physics

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Question

A car traveling at $30\frac{m}{s}$ has a kinetic energy . If the car accelerates to $35 \frac{m}{s}$ , what will the new kinetic energy be?

Answer

Kinetic energy is given by $E = \frac{1}{2}mv^2$ . We will begin by calculating the car's initial kinetic energy, in terms of the unknown mass of the car :

$E_{initial} = x = \frac{1}{2}m(30\frac{m}{s})^2$ .

Next, we will calculate the final energy of the car, also in terms of the unknown mass of the car:

$E_{final} = \frac{1}{2}m(35\frac{m}{s})^2$ .

To find the ratio of the final to initial kinetic energy, we divide $E_{final}$ by $E_{initial}$ . We see that this reduces to $\frac{35^2}{30^2}$ with the both the mass and $\frac{1}{2}$ terms cancelling. . Thus, the new kinetic energy is .

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Question

A steel ball is in a spring loaded launcher. The spring has a constant of $30\frac{N}{m}$ . If the ball is pulled back 50 cm then released and the ball leaves the launcher at a speed of $10\frac{m}{s}$ , how much work was done by friction?

Answer

So in order to solve this problem, we first should think about where all the energy in the system is coming from and going. Specifically when you pull the spring back, the work you do is turned into potential energy stored in the spring. So:

$W_{pulling spring} = PE$

and the energy transforms into kinetic when you let go, but some of that energy is lost to friction so

$PE = KE + W_{friction}$ and $PE_{spring} = \frac{1}{2} kx^2$

So use the equation below, enter in your known quantities and then solve for the work done by friction.

$\frac{1}{2} kx^2 = \frac{1}{2}mv^2 + W_{friction}$

Plug in known values and solve.

$3.75=2.5+W_{friction}$

$1.25J=W_{friction}$

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