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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
[tex]\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[/tex]