A particle A of mass 0.8 kg rests on a horizontal table and is attached to one end of a light inextensible string - Edexcel - A-Level Maths Mechanics - Question 8 - 2003 - Paper 1

Using the formula for distance covered under constant acceleration:

$s = ut + \frac{1}{2} a t^2$

Where:

• $s = 0.6 ext{ m}$ (height B falls)
• $u = 0$ (initial velocity)
• $a = g - \frac{T}{m_B}$

From previous calculation, we have found $T = 4.7 ext{ N}$, which gives

$a = 9.8 - \frac{4.7}{1.2} \approx 5.88$

Substituting into the distance formula, we can solve for $t$:

With the coefficient of friction rac{1}{3}, we need to reanalyze the forces. The frictional force $F_f$ can be calculated as:

$F_f = \mu R = \frac{1}{3} imes 0.8g$

Using this in the equations gives a new value for acceleration. Therefore, the new setup shows: $R = T + F_f$ Replace the expressions accordingly and solve: $a' = 0.52 g$ Then apply again the distance formula: $0.6 = \frac{1}{2} \cdot a' t^2$. This will yield a different time, approximately: $t \approx 0.49 ext{ s}$. Hence the time taken by B to reach the ground changes with the introduction of friction.