A block of mass 100, gm slides on a rough horizontal surface. If speed of block decreases from 10 ,m

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10-1.Thermometry, Thermal Expansion and Calorimetry

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A block of mass $100\, gm$ slides on a rough horizontal surface. If the speed of the block decreases from $10 \,m/s$ to $5\, m/s,$ the thermal energy developed in the process is............. $J$

A

$3.75$

B

$37.5$

C

$0.375$

D

$0.75$

Solution

(a) According to energy conservation, change in kinetic energy appears in the form of heat (thermal energy).

$\Rightarrow$ i.e. Thermal energy $ = \frac{1}{2}m(v_1^2 – v_2^2)$

$\left[ {\because \,\mathop W\limits_{{\text{(Joule)}}} = \mathop Q\limits_{{\text{(Joule)}}} } \right]$

$ = \frac{1}{2}(100 \times {10^{ – 3}})({10^2} – {5^2}) = 3.75\,J$

Standard 11

Physics

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