Block P, of unknown mass, is placed on a rough horizontal surface - NSC Physical Sciences - Question 2 - 2018 - Paper 1

To calculate the acceleration of the 3 kg block, we can use the second equation of motion:

$$ext{Displacement} (s) = v_i t + \frac{1}{2} a t^2$$

Given that:

• Initial velocity, $v_i = 0$ (the block starts from rest)
• Displacement, $s = 0.5 ext{ m}$
• Time, $t = 3 ext{ s}$

We substitute these values into the equation:

$$0.5 = 0 + \frac{1}{2} a (3^2)$$

Simplifying gives:

$$0.5 = \frac{1}{2} a (9)$$ $$0.5 = 4.5a$$

Thus, solving for $a$ gives:

$$a = \frac{0.5}{4.5} = 0.111 ext{ m/s}^2$$

To find the tension ($T$) in the string, we analyze the forces acting on the 3 kg block:

Using Newton's second law:

$$F_{net} = ma$$

The net force can be expressed as the weight of the block minus the tension:

$$3g - T = 3a$$

Where:

• $g = 9.81 ext{ m/s}^2$
• $a = 0.111 ext{ m/s}^2$

Substituting the values:

$$3(9.81) - T = 3(0.111)$$

Calculating gives:

$$29.43 - T = 0.333$$

Therefore:

$$T = 29.43 - 0.333 = 29.097 ext{ N} ext{ (approximately 29.1 N)}$$

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

• Weight ($W$) acting downwards due to gravity
• Tension ($T$) in the string acting horizontally towards the pulley
• Normal force ($N$) acting upwards perpendicular to the surface
• Frictional force ($f$) opposing the direction of motion

The diagram should clearly show these forces acting on the block.

To find the mass of block P, we need to use the frictional force and the equation:

f = ext{frictional force} = ext{coefficient of friction} \times N$$ We know the frictional force is given as 27 N. The normal force ($N$) acting on block P is equal to its weight:

N = mg, ext{ where } m ext{ is the mass of block P.}

Assuming a coefficient of static friction ($\mu_s$) of approximately 0.5, we can find:

27 = 0.5 \times mg

$$Rearranging gives:$$

m = \frac{27}{0.5g} = \frac{27}{0.5 \times 9.81} = 5.5 ext{ kg (approximately)}

$$$$