Question

A fireman of mass 80 kg slides down a pole. When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s. By how much has thermal energy increased during his slide?

Answer 1

3100 J

Step-by-step explanation:

Because this scenario using three forms of energy (kinetic, gravitational potential, thermal), we use the conservation of energy formula:

0=ΔKe + Δ Ug + ΔEth

Keep in mind we want to find change in thermal energy, so:

ΔEth = -(ΔKe + ΔUg)

Change in kinetic energy:

ΔKe = Kf - Ki =
`(1)/(2)(0)^(2) -(1)/(2)(80)(2.2)^(2) = 190 J`

Change in Gravitational potential energy:

`mgy_(f) - mgy_(i) = 0 - (80)(9.80)(4.2) = - 3300 J`

ΔEth =
`-(190 -3300) = 3100 J`

Change in thermal energy = 3100 J

Answer 2

3099 J

Step-by-step explanation:

While the fireman slides down, his initial gravitational potential energy is converted partially into kinetic energy, partially into thermal energy, so we can write:

`\Delta U = K + E_t` (1)

where

where m=80 kg is the mass of the fireman, g=9.8 m/s^2 is the acceleration of gravity,
`\Delta h=4.2 m` is the change in height of the fireman.

The kinetic energy gained is

`K=(1)/(2)mv^2=(1)/(2)(80 kg)(2.2 m/s)^2=194 J`

where v = 2.2 m/s is the speed reached by the fireman at the bottom of the slide

So now solving eq.(1) we find the increase in thermal energy :

`E_t = \Delta U - K = 3293 J - 194 J = 3099 J`