Answer:
4431.06 J
Explanation:
The mechanical energy is made up by potential energy and kinetic energy. Let the bottom of the stair be ground 0 for potential energy. The kinetic energy at the bottom is
[tex]E_k = mv_b^2/2 = 61.8*2^2/2 =123.6 J[/tex]
The kinetic and potential energy at the top is (let g = 9.81 m/s2
[tex] mv_t^2/2 + mgh = 61.8*0.5^2/2 + 61.8*7.5*9.81 = 4554.66 J[/tex]
So the change in mechanical energy between the top and bottom stair would be
[tex] 4554.66 - 123.6 = 4431.06J[/tex]
Answer:
Change in mechanical energy, ΔE = 4426.425 J
Explanation:
Mass of the student, m = 61.8 kg
At the bottom of the stairs:
speed, [tex]v_{B} = 2 m/s[/tex]
Mechanical Energy at the bottom of the stairs, [tex]E_{b} = \frac{1}{2} mv_{b} ^{2}[/tex]
[tex]E_{b} = \frac{1}{2} 61.8 * 2^{2}\\[/tex]
[tex]E_{b} = 123.6 J[/tex]
The potential energy [tex]E_{p}[/tex] while the boy was transiting from the bottom of the stairs to the top of the stairs.
Difference in height between top and bottom of stairs, h = 7.5m
[tex]E_{p} = mgh\\E_{p} = 61.8 * 9.8 * 7.5\\E_{p} = 4542.3 J[/tex]
At the top of stairs, [tex]v_{t} = 0.500 m/s[/tex]
Kinetic energy at the top of stairs, [tex]E_{k} = \frac{1}{2} mv_{t} ^{2}[/tex]
[tex]E_{k} = \frac{1}{2} * 61.8*0.5 ^{2}[/tex]
[tex]E_{k} = 7.725 J[/tex]
Total Mechanical energy at the top of the stairs, [tex]E_{t} = E_{k} + E_{p}[/tex]
[tex]E_{t} = 4542.3 + 7.725\\E_{t} = 4550.025 J[/tex]
Change in mechanical energy, ΔE = [tex]E_{t} - E_{b}[/tex]
ΔE = 550.025 - 123.6
ΔE = 4426.425 J
