Question

A block of steel of mass 2 kg slides down a rough inclined plane of inclination of \( \sin^{-1} \left( \frac{3}{5} \right) \) at a constant speed. The temperature of the block as it slides through 80 cm, assuming that the mechanical energy lost is used to increase the temperature of the block is nearly:

• A. \( 0.0190^\circ C \)
• B. \( 0.0114^\circ C \)
• C. \( 0.0152^\circ C \)
• D. \( 0.0952^\circ C \)

Hint:

When solving such problems, remember that all mechanical energy lost is converted into heat, and use the formula \( Q = m \cdot s \cdot \Delta T \) to find the temperature change.

Answer 1

The Correct Option is B

We know that the mechanical energy lost is converted into heat. The amount of heat required to increase the temperature of the block is given by the equation: \[ Q = m \cdot s \cdot \Delta T \] where \( m = 2 \, \text{kg} \), \( s = 420 \, \text{J/kg}^\circ C \) (specific heat capacity), and \( \Delta T \) is the change in temperature. The work done by the gravitational force is the mechanical energy lost: \[ W = m \cdot g \cdot h \] where \( g = 10 \, \text{m/s}^2 \) and \( h = 0.8 \, \text{m} \). Now, the total work done is converted into heat, so: \[ m \cdot g \cdot h = m \cdot s \cdot \Delta T \] Simplifying and solving for \( \Delta T \): \[ \Delta T = \frac{g \cdot h}{s} \] Substituting the values: \[ \Delta T = \frac{10 \cdot 0.8}{420} = 0.0114^\circ C \] Thus, the temperature increase is \( 0.0114^\circ C \).