Answer:
option (c)
Explanation:
h1 = 26 m
h2 = 16 m
Let the speed of the car is v.
Use the law of energy conservation
Potential energy at initial point = Potential energy at final point + Kinetic energy at final point
m x g x h1 = m x g x h2 + 1/2 x m x v^2
9.8 x 26 - 9.8 x 16 = 0.5 x v^2
196 = v^2
v = 14 m/s
The speed of the car at the top of the hill is C) 14 m/s
[tex]\texttt{ }[/tex]
Let's recall the formula of Kinetic Energy as follows:
[tex]\large {\boxed {E_k = \frac{1}{2}mv^2 }[/tex]
Ek = Kinetic Energy ( Newton )
m = Object's Mass ( kg )
v = Speed of Object ( m/s )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
initial height = h₁ = 26 m
final height = h₂ = 16 m
initial speed = v₁ = 0 m/s
Asked:
final speed = v₂ = ?
Solution:
We will use Conservation of Energy formula to solve this problem as follows:
[tex]Ep_1 + Ek_1 = Ep_2 + Ek_2[/tex]
[tex]mgh_1 + \frac{1}{2}m(v_1)^2 = mgh_2 + \frac{1}{2}m(v_2)^2[/tex]
[tex]mgh_1 + \frac{1}{2}m(0)^2 = mgh_2 + \frac{1}{2}m(v_2)^2[/tex]
[tex]mgh_1 = mgh_2 + \frac{1}{2}m(v_2)^2[/tex]
[tex]gh_1 = gh_2 + \frac{1}{2}(v_2)^2[/tex]
[tex]2g(h_1 - h_2) = (v_2)^2[/tex]
[tex]v_2 = \sqrt {2g(h_1 - h_2)}[/tex]
[tex]v_2 = \sqrt { 2(9.8)(26 - 16) }[/tex]
[tex]v_2 = \sqrt { 196 }[/tex]
[tex]v_2 = 14 \texttt{ m/s}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Dynamics
