Question

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

Complete 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:

The thermal energy increased by 3,099.2 J

Explanation:

Given;

mass of the fireman, m = 80 kg

initial position of the fireman, hi = 4.2 m

final speed, v = 2.2 m/s

The change in the thermal energy is calculated as;

ΔE +  (K.Ef - K.Ei) + (Uf - Ui) = 0

where;

ΔE is the change in the thermal energy

K.Ef is the final kinetic energy

K.Ei is the initial kinetic energy

Uf is the final potential energy

Ui is the initial potential energy

$\Delta E_{th} + (\frac{1}{2} mv_f^2 - \frac{1}{2} mv_i^2) + (mgh_f - mgh_i)=0\\\\initial \ velocity, \ v_i = 0\\final \ height , \ h_f = o\\\\\Delta E_{th} + (\frac{1}{2} mv_f^2) + ( - mgh_i)=0\\\\\Delta E_{th} + \frac{1}{2} mv_f^2 - mgh_i = 0\\\\\Delta E_{th} =mgh_i - \frac{1}{2} mv_f^2\\\\\Delta E_{th} = 80 \times 9.8 \times 4.2 \ \ - \ \ \frac{1}{2} \times 80 \times (2.2)^2 \\\\\Delta E_{th} = 3292.8 \ J \ - \ 193.6 \ J\\\\\Delta E_{th} = 3,099.2 \ J$