here we can use the concept of mechanical energy conservation
[tex]U_i + KE_i = U_f + KE_f[/tex]
[tex]mgh_1 + \frac{1}{2}mv_i^2 = mgh_2 + \frac{1}{2}mv_f^2[/tex]
now we know that
[tex]v_i = 4 m/s[/tex]
m = 60 kg
[tex]h_1 = 0 m[/tex]
[tex]h_2 = - 10 m[/tex] As its going down
now from above equation we will have
[tex]0 + \frac{1}{2}(60)(4^2) = (60)(9.81)(-10) + \frac{1}{2}(60)v_f^2[/tex]
[tex]480 = -5886 + 30v_f^2[/tex]
[tex]v_f^2 = 212.2[/tex]
[tex]v_f = 14.57 m/s[/tex]
so his speed will be 14.57 m/s
She will be moving as fast as 15 m/s after dropping 10 m in elevation
[tex]\texttt{ }[/tex]
Let's recall the Kinetic Energy formula:
[tex]\boxed {E_k = \frac{1}{2}mv^2 }[/tex]
Ek = kinetic energy ( J )
m = mass of object ( kg )
v = speed of object ( m/s )
[tex]\texttt{ }[/tex]
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration (m / s²)v = final velocity (m / s)
u = initial velocity (m / s)
t = time taken (s)
d = distance (m)
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
gravitational acceleration = g = 9.8 m/s²
initial height = h_1 = 10 m
final heght = h_2 = 0 m
initial speed = v_1 = 4.0 m/s
Asked:
final speed = v_2 = ?
Solution:
We will calculate the speed of the skier by using Conservation of Energy formula 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]60(9.8)(10) + \frac{1}{2}(60)(4.0)^2 = 60(9.8)(0) + \frac{1}{2}(60)(v_2)^2[/tex]
[tex]6360 = 30(v_2)^2[/tex]
[tex](v_2)^2 =212[/tex]
[tex]v_2 = \sqrt{212}[/tex]
[tex]v_2 \approx 15 \texttt{ m/s}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Mathematics
Chapter: Energy
