Analysis of Statement 1

Statement 1: 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° 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.

1. Define Decrease in Mechanical Energy: The decrease in mechanical energy of a system is equal to the work done by non-conservative forces acting on it. In this problem, the only non-conservative force is friction. The work done by friction ($W_f$) is always negative as it opposes motion. Therefore, the decrease in mechanical energy is $|W_f|$. Alternatively, decrease in mechanical energy = $E_{initial} - E_{final}$.

2. Situation 1: Horizontal Surface

   • The block starts with initial velocity $v$ and comes to rest.
   • Let the horizontal surface be the reference level for potential energy ($U=0$).
   • Initial mechanical energy, $E_{i1} = K_{i1} + U_{i1} = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2$.
   • Final mechanical energy, $E_{f1} = K_{f1} + U_{f1} = 0 + 0 = 0$.
   • The decrease in mechanical energy in the first situation is $D_1 = E_{i1} - E_{f1} = \frac{1}{2}mv^2$. This energy is completely converted into heat due to friction.

3. Situation 2: Inclined Surface (going up)

   • The block starts with the same initial velocity $v$ at the bottom of the incline.
   • Let the starting point be the reference level for potential energy ($U=0$).
   • Initial mechanical energy, $E_{i2} = K_{i2} + U_{i2} = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2$.
   • The block moves up the incline, say by a vertical height $h$, and comes to rest. At its highest point, its kinetic energy is zero.
   • Final mechanical energy, $E_{f2} = K_{f2} + U_{f2} = 0 + mgh = mgh$.
   • The decrease in mechanical energy in the second situation is $D_2 = E_{i2} - E_{f2} = \frac{1}{2}mv^2 - mgh$.
   • From the work-energy theorem for non-conservative systems, the change in mechanical energy is equal to the work done by friction: $\Delta E = E_{f2} - E_{i2} = W_f$. So, $mgh - \frac{1}{2}mv^2 = W_f$. The energy dissipated by friction is $|W_f| = \frac{1}{2}mv^2 - mgh$. This is also the decrease in mechanical energy, $D_2$.

4. Comparison

   • $D_1 = \frac{1}{2}mv^2$.
   • $D_2 = \frac{1}{2}mv^2 - mgh$.
   • Since the block moves up the incline before stopping, the final height $h$ is greater than zero ($h>0$).
   • Therefore, $mgh > 0$.
   • This implies that $D_2 = \frac{1}{2}mv^2 - mgh < \frac{1}{2}mv^2 = D_1$.
   • So, the decrease in mechanical energy in the second situation is smaller than in the first situation.
   • Statement 1 is True.

Analysis of Statement 2

Statement 2: The coefficient of friction between the block and the surface decreases with the increase in the angle of inclination.

1. Definition of Coefficient of Friction: The coefficient of friction ($\\\mu$) is a dimensionless quantity that characterizes the frictional properties between two surfaces in contact. In the standard model of friction (Amontons' laws), the coefficient of friction is assumed to be a constant that depends only on the nature of the two surfaces.

2. Dependence on Angle: The angle of inclination, $\ heta$, affects the normal force ($N$) between the surfaces. On an incline, $N = mg \cos\theta$. As $\ heta$ increases, $\cos\theta$ decreases, so the normal force $N$ decreases. This, in turn, decreases the frictional force, $f = \mu N = \mu mg \cos\theta$.

3. Conclusion: While the frictional force changes with the angle of inclination, the coefficient of friction, $\\\mu$, is considered to be a constant property of the surfaces and does not depend on the angle.

   • Statement 2 is False.

Final Conclusion

• Statement 1 is True.
• Statement 2 is False.

This corresponds to option C.