Question

A 60 kg student slides down a 5.0 metre high slide. At the bottom of the slide , the student is moving at 4.5 m/s (due to friction) . If the slide is a 10.0 Kg layer of steel , by how much will the steel temperature increase?

Answer 1

Energy lost by the ball will be used in increasing the temperature.

i.e. $\Delta E = \Delta Q$

$m_1gh-\frac{1}{2}m_1v^2 = m_ss\Delta T$

where v is the final velocity of student

h is the height of student

s is the specific heat of steel

$m_1$ is the mass of student

$m_s$ is the mass of steel

$\Delta T$ is the temperature change

Putting values,

$(60 \times 9.8 \times 5) - (\frac{1}{2} \times 60 \times 4.5^2) = 10 \times 420 \times \Delta T$

Solving it we get,

$\Delta T = 0.55 ^\circ C$