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

Change in mechanical energy, ΔE = 4426.425 J

Step-by-step 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) = (1)/(2) mv_(b) ^(2)`

`E_(b) = (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) = (1)/(2) mv_(t) ^(2)`

`E_(k) = (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

Answer 2

4431.06 J

Step-by-step 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`