Question

the box that bob and jane are in ( total \\(m = 200\text{ kg}\\) ) is sitting precariously close to a ramp that leads down to the basement. just as jane finishes her sentence, the box is pushed and slides down the ramp! the box starts at the top of the frictionless ramp shown below with a velocity of \\(v_0 = 5.00\text{ m/s}\\).

how fast is the box moving at the bottom of the frictionless ramp in meters per second?

question 2

at the bottom of the ramp is a bit of floor where the coefficient of kinetic friction is 0.580. starting from the bottom of the ramp, what is the distance that the box travels before it comes to a stop?

Explanation:

Identify given parameters

We extract the parameters:

• Mass of the box \(m = 200\text{ kg}\)
• Initial velocity at the top of the ramp \(v_0 = 5.00\text{ m/s}\)
• Height of the ramp \(h = 7\text{ m}\)
• Acceleration due to gravity \(g = 9.8\text{ m/s}^2\)
• Coefficient of kinetic friction on the flat floor \(\mu_k = 0.580\) (noting the handwritten \(0.587\) on the diagram, but the printed text in Question 2 specifies \(0.580\); we will use \(0.580\) as per the printed text of Question 2)

Calculate velocity at the bottom

Using the conservation of energy:

$$ E_i = E_f $$

$$ \frac{1}{2}mv_0^2 + mgh = \frac{1}{2}mv_f^2 $$

Divide by \(m\):

$$ \frac{1}{2}v_0^2 + gh = \frac{1}{2}v_f^2 $$

$$ v_f = \sqrt{v_0^2 + 2gh} $$

Substitute the values:

$$ v_f = \sqrt{5.00^2 + 2 \times 9.8 \times 7} = \sqrt{25 + 137.2} = \sqrt{162.2} \approx 12.74\text{ m/s} $$

Calculate stopping distance

Using the work-energy theorem on the flat floor:

$$ W_{\text{friction}} = \Delta K $$

$$ -f_k \cdot s = 0 - \frac{1}{2}mv_f^2 $$

Since \(f_k = \mu_k F_N = \mu_k mg\):

$$ -\mu_k mg \cdot s = -\frac{1}{2}mv_f^2 $$

Divide by \(-mg\):

$$ s = \frac{v_f^2}{2\mu_k g} $$

Substitute \(v_f^2 = 162.2\) and \(\mu_k = 0.580\):

$$ s = \frac{162.2}{2 \times 0.580 \times 9.8} = \frac{162.2}{11.368} \approx 14.27\text{ m} $$

Answer:

Question 1

\(12.74\)

Question 2

\(14.27\)