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प्रश्न
A copper cube of mass 200 g slides down on a rough inclined plane of inclination 37° at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = 420 J kg−1 K−1.
उत्तर
Mass of copper cube, m = 200 g = 0.2 kg
Length through which the block has slided, l = 60 cm = 0.6 m
Since the block is moving with constant velocity, the net force on it is zero. Thus,
Force of friction, f = mg
Also, since the object is moving with a constant velocity, change in its K.E will be zero.As the object slides down, its PE decreases at the cost of increase in thermal energy of copper.
The loss in mechanical energy of the copper block = Work done by the frictional force on the copper block to a distanceof 60 cm
W = mg l sin θ
W = 0.2 × 10 × 0.6 sin 37°
`W=1.2xx(3/5)=0.72`
Let the change in temperature of the block be ∆T.
Thermal energy gained by block = ms ∆T = 0.2 × 420×∆T = 84∆T
But 84∆T = 0.72
`rArrDeltaT=0.72/84=0.00857`
`Delta=0.0086=8.6xx10^-3°C`
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