JEE Advanced · Physics · 6. Work Power Energy
Statement I A block of mass \(m\) starts moving on a rough horizontal surface with a velocity \(v\). It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of \(30^{\circ}\) with the horizontal and the same block is made to go up on the surface with the same initial velocity \(v\). The decrease in the mechanical energy in the second situation is smaller than that in the first situation.
Statement II The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.
- A Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I
- B Statement I is true, Statement II is true; Statement II is NOT a correct explanation for Statement I
- C Statement I is true, Statement II is false
- D Statement I is false, Statement II is true
Answer & Solution
Correct Answer
(C) Statement I is true, Statement II is false
Step-by-step Solution
Detailed explanation
In statement I Decrease in mechanical energy in case I will be
\(
\Delta U_1=\frac{1}{2} m v^2
\)
But decrease in mechanical energy in case II will be
\(
\begin{aligned}
& \Delta U_2=\frac{1}{2} m v^2-m g h \\
& \Delta U_2 < \Delta U_1
\end{aligned}
\)
\(
\therefore \quad \Delta U_2 < \Delta U_1
\)
or Statement \(\mathrm{I}\) is correct.
In Statement II Coefficient of friction will not change or this Statement is wrong.
\(\therefore\) Option (c) is correct.
\(
\Delta U_1=\frac{1}{2} m v^2
\)
But decrease in mechanical energy in case II will be
\(
\begin{aligned}
& \Delta U_2=\frac{1}{2} m v^2-m g h \\
& \Delta U_2 < \Delta U_1
\end{aligned}
\)
\(
\therefore \quad \Delta U_2 < \Delta U_1
\)
or Statement \(\mathrm{I}\) is correct.
In Statement II Coefficient of friction will not change or this Statement is wrong.
\(\therefore\) Option (c) is correct.
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