An 8-kg block, on a rough horizontal surface, is connected to a 2-kg block by a light inextensible string passing over a frictionless pulley, as shown below - NSC Physical Sciences - Question 2 - 2021 - Paper 1

The free-body diagram for the 8-kg block should include the following forces:

• The gravitational force ($F_g = mg = 8 imes 9.8 = 78.4 ext{ N}$) acting downwards.
• The normal force ($F_N = 78.4 ext{ N}$) acting upwards.
• The applied force ($F = 29.6 ext{ N}$) acting horizontally to the right.
• The frictional force ($f_f = 10 ext{ N}$) acting horizontally to the left.

Label these forces appropriately in the diagram.

To find the tension in the string, we can analyze the forces acting on the 2-kg block. Since the system is moving at a constant speed, the net force on the 2-kg block is zero.

The forces acting on the 2-kg block are:

• Tension ($T$) acting upward.
• Weight ($W = mg = 2 imes 9.8 = 19.6 ext{ N}$) acting downward.

Setting up the equilibrium equation:

ightarrow T = W ightarrow T = 19.6 ext{ N}$$

For the 8-kg block, the net force can be expressed as:

$F_{ ext{net}} = F_{ ext{applied}} - f_f = 29.6 - 10 = 19.6 ext{ N}$

Applying Newton's Second Law:

ightarrow 19.6 = 8a$$ Solving for acceleration ($a$): $$a = rac{19.6}{8} = 2.45 ext{ m/s}^2$$

As previously calculated in step 2.3, the tension in the string is:

$T = 19.6 ext{ N}$