Question

5.7 systems and energy conservation

1. a 50.-kg pole vaulter running at 10. m/s vaults over the bar. her speed when she is above the bar is 1.0 m/s. neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.
2. a child and a sled with a combined mass of 50.0 kg slide down a frictionless slope. if the sled starts from rest and has a speed of 3.00 m/s at the bottom, what is the height of the hill?

Explanation:

Problem 32

Step1: Identify Energy Conservation

Use conservation of mechanical energy: Initial kinetic energy ($K_i$) + Initial potential energy ($U_i$) = Final kinetic energy ($K_f$) + Final potential energy ($U_f$). Assume initial height $h_i = 0$, so $U_i = 0$.
$K_i = \frac{1}{2}mv_i^2$, $K_f = \frac{1}{2}mv_f^2$, $U_f = mgh_f$.

Step2: Set Up Equation

$\frac{1}{2}mv_i^2 + 0 = \frac{1}{2}mv_f^2 + mgh_f$. Cancel $m$: $\frac{1}{2}v_i^2 = \frac{1}{2}v_f^2 + gh_f$.

Step3: Solve for $h_f$

Rearrange: $h_f = \frac{v_i^2 - v_f^2}{2g}$. Plug $v_i = 10\,\text{m/s}$, $v_f = 1\,\text{m/s}$, $g = 9.8\,\text{m/s}^2$.
$h_f = \frac{10^2 - 1^2}{2\times9.8} = \frac{100 - 1}{19.6} = \frac{99}{19.6} \approx 5.05\,\text{m}$.

Step1: Apply Energy Conservation

Initial energy: $K_i + U_i = 0 + mgh$ (starts from rest, $v_i = 0$, initial height $h$). Final energy: $K_f + U_f = \frac{1}{2}mv_f^2 + 0$ (at bottom, $h_f = 0$).

Step2: Equate Energies

$mgh = \frac{1}{2}mv_f^2$. Cancel $m$: $gh = \frac{1}{2}v_f^2$.

Step3: Solve for $h$

$h = \frac{v_f^2}{2g}$. Plug $v_f = 3.00\,\text{m/s}$, $g = 9.8\,\text{m/s}^2$.
$h = \frac{3.00^2}{2\times9.8} = \frac{9.00}{19.6} \approx 0.459\,\text{m}$.

Answer:

$\approx 5.1\,\text{m}$ (or $5.05\,\text{m}$)

Problem 33