Question

1. the height h of the highest hill in the figure is 15 m. what would have to be the speed of a sled at the top of that hill (at point a) in order that it arrive at point d with a speed of 24 m/s? ignore friction.

(a) 12.3
(b) 14.6
(c) 8.6
(d) 19.5
(e) none of these

Explanation:

Step1: Apply conservation of mechanical energy

The conservation - of - mechanical - energy formula is $E_{i}=E_{f}$, where $E = K+U$. Kinetic energy $K=\frac{1}{2}mv^{2}$ and potential energy $U = mgh$. Let the speed at point A be $v_{A}$ and at point D be $v_{D}=24\ m/s$. The height at point A is $h_{A}=15\ m$ and at point D is $h_{D}=5\ m$. So, $\frac{1}{2}mv_{A}^{2}+mgh_{A}=\frac{1}{2}mv_{D}^{2}+mgh_{D}$.

Step2: Cancel out the mass m

Since $m$ is non - zero, we can divide the entire equation $\frac{1}{2}mv_{A}^{2}+mgh_{A}=\frac{1}{2}mv_{D}^{2}+mgh_{D}$ by $m$ to get $\frac{1}{2}v_{A}^{2}+gh_{A}=\frac{1}{2}v_{D}^{2}+gh_{D}$.

Step3: Rearrange the equation to solve for $v_{A}$

First, multiply through by 2 to get $v_{A}^{2}+2gh_{A}=v_{D}^{2}+2gh_{D}$. Then, $v_{A}^{2}=v_{D}^{2}+2g(h_{D}-h_{A})$.

Step4: Substitute the given values

Given $g = 9.8\ m/s^{2}$, $v_{D}=24\ m/s$, $h_{A}=15\ m$, and $h_{D}=5\ m$. Substitute these values into the equation: $v_{A}^{2}=(24)^{2}+2\times9.8\times(5 - 15)$.
$v_{A}^{2}=576+19.6\times(- 10)$.
$v_{A}^{2}=576-196$.
$v_{A}^{2}=380$.
$v_{A}=\sqrt{380}\approx19.5\ m/s$.

Answer:

(d) 19.5