# Physics Word Problem?

## Jun 29, 2016

$4.2 m$

#### Explanation:

Step 1
According to the law of conservation of energy, the total initial energy is equal to the total final energy. That is, the total energy at X is the same at Y. So we know that

${E}_{T , i} = {E}_{T , f}$

If we simplify the equation further, we get

${E}_{k , i} + {E}_{g , i} = {E}_{k , f} + {E}_{g , f}$

Note: The distance between X and Y has no friction, so it is not included in the equation!

$\frac{1}{2} m {v}_{i}^{2} + m g {h}_{i} = \frac{1}{2} m {v}_{f}^{2} + m g {h}_{f}$

Since we already know the initial velocity and height, as well as the final height, the only variable left to solve for is the final velocity at Y. Knowing the final velocity will help find how far the cart moves past Y. Thus, solve for ${v}_{f}$,

$\textcolor{red}{\cancel{\textcolor{b l a c k}{m}}} \left(\frac{1}{2} {v}_{i}^{2} + g {h}_{i}\right) = \textcolor{red}{\cancel{\textcolor{b l a c k}{m}}} \left(\frac{1}{2} {v}_{f}^{2} + g {h}_{f}\right)$

$\frac{1}{2} {v}_{f}^{2} = \frac{1}{2} {v}_{i}^{2} + g {h}_{i} - g {h}_{f}$

${v}_{f} = \sqrt{\frac{\frac{1}{2} {v}_{i}^{2} + g \left({h}_{i} - {h}_{f}\right)}{\frac{1}{2}}}$

Plugging in the values,

$\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{{v}_{f}} = \sqrt{\frac{\frac{1}{2} {\left(12 \frac{m}{s}\right)}^{2} + \left(9.81 \frac{m}{s} ^ 2\right) \left(4.0 m - 6.0 m\right)}{\frac{1}{2}}}$

$= \textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{10.24 \frac{m}{s}}$

Step 2
For this section, we will let the up and right directions be positive.

In order to determine how far the cart moves past Y, we must also determine the acceleration. This can be found with Newton's second law,

${F}_{\text{net}} = m a$

If we sum all the forces acting on the cart, the equation becomes

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{F}_{N}}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{{F}_{g}}}} + {F}_{f} = m a$

Note: The normal and gravity forces cancel each other out since the cart does not accelerate in the vertical direction!

${F}_{f} = m a$

Solving for the acceleration of the cart,

$\textcolor{p u r p \le}{a} = {F}_{f} / m$

$= \frac{- 25 N}{2.0 k g}$

Note: The force of friction is negative because we let the right direction be positive and friction acts against that direction in which the cart is moving!

$= \textcolor{p u r p \le}{- 12.5 \frac{m}{s} ^ 2} \textcolor{w h i t e}{X X}$ (the negative sign means the cart is not $\textcolor{w h i t e}{X X X X X X X X X \times}$accelerating, but decelerating!)

Step 3
After determining the initial velocity and acceleration of the cart as it starts to slow down at Y, we have enough information to use the following formula to determine how far it will move until it slows at a halt:

${v}_{f}^{2} = {v}_{i}^{2} + 2 a \Delta d$

Solving for $\Delta d$,

$\Delta d = \frac{{v}_{f}^{2} - {v}_{i}^{2}}{2 a}$

Plugging in the values,

$= \frac{{\left(0 \frac{m}{s}\right)}^{2} - {\left(\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{10.24 \frac{m}{s}}\right)}^{2}}{2 \left(\textcolor{p u r p \le}{- 12.5 \frac{m}{s} ^ 2}\right)}$

$= 4.19 m$

$\approx \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{4.2 m} \textcolor{w h i t e}{\frac{a}{a}} |}}}$