A small stone A of mass 3m is attached to one end of a string - Edexcel - A-Level Maths Mechanics - Question 2 - 2021 - Paper 1

To find the acceleration, we can analyze the forces further. The total forces perpendicular to the incline can be resolved:

1. Sum of forces acting on A:
   • Normal force ( R = 3mg \cos(\alpha) )
   • Given: ( \mu = \frac{1}{6}), the frictional force is (F = \mu R = \frac{1}{6} (3mg \cos(\alpha)))

With ( \cos(\alpha) = \frac{4}{5}):
[ F = \frac{1}{6}(3mg \cdot \frac{4}{5}) = \frac{2mg}{5} ]

2. Inserting back into the equation:

[ 3mg \frac{3}{5} - T - \frac{2mg}{5} = 3a ]

3. Rearranging yields:\n [ T = 3mg (\frac{3}{5}) - \frac{2mg}{5} ] [ T = \frac{9mg}{5} - \frac{2mg}{5} = \frac{7mg}{5} ]

Now place this back in the motion equation to solve for acceleration: [ 3mg \cdot \frac{3}{5} - \frac{7mg}{5} = 3a ] This gives: [ \frac{9mg}{5} - \frac{7mg}{5} = 3a ] [ \frac{2mg}{5} = 3a ]
From this, we see: [ a = \frac{2}{15}g ]

This graph reflects B’s velocity increasing linearly with respect to time until B reaches the pulley.

The initial conditions and the acceleration calculated in part (b) depend on the tension experienced by both masses. Given that stone B is now experiencing acceleration, it influences the forces acting on stone A and may result in a different tension in the string than initially presumed.

If the masses or configuration of the system changes, the frictional effects and acceleration could differ, thereby affecting the total result derived in part (b) regarding A's motion. If the velocity of B increases due to heavier loading or friction anomalies, we must reassess our calculations accordingly.