Question

A student approaches the outdoor staircase near Hope Commons, and takes a deep breath before attempting to climb all 30 stairs. At the bottom of the stairs, the student (61.8 kg) is moving at 2.00 m/s; at the top of the stairs the student is tired and only moving at 0.500 m/s. The top of the stairs is 7.50 meters higher than the bottom of the stairs. What is the change in mechanical energy

Answer 1

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

$E_k = mv_b^2/2 = 61.8*2^2/2 =123.6 J$

The kinetic and potential energy at the top is (let g = 9.81 m/s2

$mv_t^2/2 + mgh = 61.8*0.5^2/2 + 61.8*7.5*9.81 = 4554.66 J$

So the change in mechanical energy between the top and bottom stair would be

$4554.66 - 123.6 = 4431.06J$

Answer 2

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, $v_{B} = 2 m/s$

Mechanical Energy at the bottom of the stairs, $E_{b} = \frac{1}{2} mv_{b} ^{2}$

$E_{b} = \frac{1}{2} 61.8 * 2^{2}$

$E_{b} = 123.6 J$

The potential energy $E_{p}$ 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

$E_{p} = mgh\\E_{p} = 61.8 * 9.8 * 7.5\\E_{p} = 4542.3 J$

At the top of stairs, $v_{t} = 0.500 m/s$

Kinetic energy at the top of stairs, $E_{k} = \frac{1}{2} mv_{t} ^{2}$

$E_{k} = \frac{1}{2} * 61.8*0.5 ^{2}$

$E_{k} = 7.725 J$

Total Mechanical energy at the top of the stairs, $E_{t} = E_{k} + E_{p}$

$E_{t} = 4542.3 + 7.725\\E_{t} = 4550.025 J$

Change in mechanical energy, ΔE = $E_{t} - E_{b}$

ΔE = 550.025 - 123.6

ΔE = 4426.425 J