Analysis of Statement 1

STATEMENT 1 : It is easier to pull a heavy object than to push it on a level ground.

To determine if this statement is true, we need to analyze the forces involved in both pushing and pulling an object. Let's assume a force $F$ is applied at an angle $\theta$ to the horizontal.

1. Case 1: Pushing the object

   • A force $F$ is applied at an angle $\theta$ below the horizontal.
   • We resolve the force $F$ into its horizontal and vertical components:
     • Horizontal component: $F_{x} = F \cos\theta$ (responsible for motion)
     • Vertical component: $F_{y} = F \sin\theta$ (acts downwards, adding to the weight)
   • The free-body diagram shows the forces: weight ($mg$) downwards, vertical component of push ($F\sin\theta$) downwards, normal force ($N_{push}$) upwards, and frictional force ($f_{push}$) opposing motion.
   • For vertical equilibrium, the net vertical force is zero: $N_{push} - mg - F \sin\theta = 0$ $N_{push} = mg + F \sin\theta$
   • The frictional force is proportional to the normal force: $f_{push} = \mu N_{push}$, where $\mu$ is the coefficient of friction. $f_{push} = \mu (mg + F \sin\theta)$
   • To move the object, the horizontal component of the applied force must overcome the frictional force: $F \cos\theta \geq f_{push}$.

2. Case 2: Pulling the object

   • A force $F$ is applied at an angle $\theta$ above the horizontal.
   • We resolve the force $F$ into its components:
     • Horizontal component: $F_{x} = F \cos\theta$
     • Vertical component: $F_{y} = F \sin\theta$ (acts upwards, opposing the weight)
   • The free-body diagram shows the forces: weight ($mg$) downwards, vertical component of pull ($F\sin\theta$) upwards, normal force ($N_{pull}$) upwards, and frictional force ($f_{pull}$) opposing motion.
   • For vertical equilibrium: $N_{pull} + F \sin\theta - mg = 0$ $N_{pull} = mg - F \sin\theta$
   • The frictional force is: $f_{pull} = \mu N_{pull} = \mu (mg - F \sin\theta)$
   • To move the object, $F \cos\theta \geq f_{pull}$.

3. Comparison

   • Comparing the normal forces, we see that $N_{push} = mg + F \sin\theta$ and $N_{pull} = mg - F \sin\theta$. Clearly, $N_{push} > N_{pull}$.
   • Since the frictional force is directly proportional to the normal force ($f = \mu N$), it follows that the frictional force is greater when pushing than when pulling: $f_{push} > f_{pull}$.
   • Therefore, a smaller applied force $F$ is required to overcome friction when pulling the object compared to pushing it. This makes pulling easier.
   • Conclusion: Statement 1 is TRUE.

Analysis of Statement 2

STATEMENT 2 : The magnitude of frictional force depends on the nature of the two surfaces in contact.

1. The formula for the maximum static friction is $f_{s,max} = \mu_s N$ and for kinetic friction is $f_k = \mu_k N$.
2. The coefficients of static friction ($\mu_s$) and kinetic friction ($\mu_k$) are empirical constants that depend on the properties of the two surfaces in contact, such as their material composition and roughness.
3. For example, the coefficient of friction between wood and concrete is different from that between rubber and asphalt. This directly affects the magnitude of the frictional force for a given normal force.
4. Conclusion: Statement 2 is TRUE.

Relationship between Statement 1 and Statement 2

• Statement 1 is true because the normal force changes depending on whether you push or pull. When pulling, the upward component of the force reduces the normal force, thus reducing friction. When pushing, the downward component increases the normal force, thus increasing friction.
• Statement 2 is a general principle of friction, explaining that the coefficient of friction $\mu$ is determined by the surfaces.
• In the scenario described in Statement 1, the object and the ground remain the same, so the nature of the surfaces and the coefficient of friction $\mu$ are constant for both pushing and pulling.
• The reason pulling is easier than pushing is due to the change in the normal force, not a change in the nature of the surfaces.
• Therefore, while Statement 2 is a true statement about friction, it does not provide the correct explanation for Statement 1.

Final Decision

• Statement 1 is True.
• Statement 2 is True.
• Statement 2 is NOT the correct explanation for Statement 1.

This corresponds to option B.