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Question: A copper cube of \[200g\] slides down on a rough inclined plane of inclination \(37^\circ \) at a co...

A copper cube of 200g200g slides down on a rough inclined plane of inclination 37∘37^\circ at a constant speed. Assume mechanical energy goes into the copper block as thermal energy. Find the increase in temperature of the block as it slides down through 60cm. Specific heat capacity of copper is 420Jkg−1K−1420Jk{g^{ - 1}}{K^{ - 1}}.
A. 0.860C{0.86^0}C
B. 0.0860C{0.086^0}C
C. 0.00860C{0.0086^0}C
D.None

Explanation

Solution

Mechanical energy is the sum of the kinetic and potential energy of the system. For a conservative system, mechanical energy is constant. That is, the change in kinetic energy is balanced by the change in potential energy. The work-energy theorem gives the relation between work and mechanical energy.

Complete step by step solution:
Let us first write the given information in the question.
Copper cube mass =200g=0.2kg = 200g = 0.2kg , inclination=37o{\text{inclination}} = {37^o}, speed=constant, mechanical energy goes into thermal energy, block slides =60cm=0.6m = 60cm = 0.6m , the specific capacity of copper = 420J/kg−K420J/kg - K.
We have to find the increase in temperature of the block as it slides down.
Now, here the copper cube is sliding with the constant speed it means the acceleration will be zero, as a result, the net force on the block will be zero. Also, due to the constant speed, there will be no change in the kinetic energy of the copper cube. So, the potential energy will decrease to counterbalance the frictional force.
Therefore, we can write,
Decrease in mechanical energy = work done by friction during sliding the block 60cm.
Therefore, we can write the following.
W=mghvertical=mghsin⁡θW = mg{h_{vertical}} = mgh\sin \theta
Let us substitute the values.
W=0.2×10×0.6×sin⁡37=1.2×35=0.72J.........(1)W = 0.2 \times 10 \times 0.6 \times \sin 37 = 1.2 \times \dfrac{3}{5} = 0.72J.........(1)
If the increase in temperature is ΔT\Delta Tthen thermal energy of the copper cube = mCΔTmC\Delta T
The gain in thermal energy =0.2×420×ΔT.........(2) = 0.2 \times 420 \times \Delta T.........(2)
Equating equations (1) and (2).
0.2×420×ΔT=0.72⇒ΔT=8.57×10−3∼0.0086oC0.2 \times 420 \times \Delta T = 0.72 \Rightarrow \Delta T = 8.57 \times {10^{ - 3}} \sim {0.0086^o}C
Hence, the correct option is (C) 0.0086oC{0.0086^o}C

Note:
The friction force is a non-conservative force, which acts between the relative motion of two rough surfaces.
Specific heat capacity is the heat required to increase the temperature of the body by one degree Celsius.