Let's address the questions from the image.

4) a) State Newton's Laws.

1. First Law (Law of Inertia): An object remains at rest or in uniform motion unless acted upon by an external force.
2. Second Law (Law of Acceleration): The force acting on an object is equal to the mass of the object times its acceleration, $F = ma$.
3. Third Law (Action and Reaction): For every action, there is an equal and opposite reaction.

4) b) Analyze the System

• Object A (10 kg) is on an inclined plane with an angle of $60^\circ$.
• Object B (25 kg) is hanging vertically connected via a pulley system.

We need to calculate forces and tensions and analyze movement.

4) c) Indicate Forces

• For Object A (on the incline):

  • The weight force $W_A = 10 \times 9.8 = 98 \, \text{N}$, which has two components:
    • $W_A \sin(60^\circ)$ down the incline.
    • $W_A \cos(60^\circ)$ perpendicular to the incline.
  • Tension force $T$ in the rope (upward along the incline).
  • Normal force $N$ perpendicular to the incline.
  • Frictional force (if there’s friction, it opposes motion).

• For Object B:

  • Weight force $W_B = 25 \times 9.8 = 245 \, \text{N}$ downwards.
  • Tension force $T$ in the rope (upward).

4) d) Calculate the Tension (No Movement)

If there is no movement, the forces are in equilibrium.

• Object B: $T = W_B = 245 \, \text{N} \quad \text{(as B is stationary, tension balances weight)}$

• Object A: The force down the incline due to weight is $W_A \sin(60^\circ) = 98 \times 0.866 = 84.87 \, \text{N}$. For no movement, $T = 84.87 \, \text{N}$, so $T$ must balance the component of weight along the incline.

Thus, $T = 84.87 \, \text{N}$.

4) e) If Object B Moves Down with $3 \, \text{m/s}^2$

• For Object B: Using Newton’s Second Law, $F = ma$: $W_B - T = m_B a = 25 \times 3 = 75 \, \text{N}$ $245 - T = 75 \quad \Rightarrow \quad T = 245 - 75 = 170 \, \text{N}$

• For Object A: Net force on A along the incline: $T - W_A \sin(60^\circ) - f = m_A a = 10 \times 3 = 30 \, \text{N}$ Substituting known values: $170 - 84.87 - f = 30 \quad \Rightarrow \quad f = 170 - 84.87 - 30 = 55.13 \, \text{N}$ So, the frictional force $f = 55.13 \, \text{N}$.

Would you like more details or have any specific questions on any part? Here are some related questions for further study:

1. How would friction affect the equilibrium in part (d)?
2. What happens to the system if the pulley is no longer frictionless?
3. How would different inclines (angles) impact the tension in the rope?
4. How does the normal force change with increasing angle of incline?
5. How can we calculate the coefficient of friction between Object A and the incline?

Tip: In inclined plane problems, always break forces into components parallel and perpendicular to the incline for easier analysis.