4. (a) Two particles of masses 3 kg and 2 kg are connected by a taut, light, inextensible string which passes over a smooth light pulley at the edge of a smooth horizontal table - Leaving Cert Applied Maths - Question 4 - 2009

Using Newton's second law:

For the 3 kg mass:

[ T = 3a ]

For the 2 kg mass:

[ 2g - T = 2a ]

Substituting the first equation into the second:

[ 2g - 3a = 2a ]

Rearranging gives:

[ 5a = 2g ]

Thus, the common acceleration is:

[ a = \frac{2g}{5} = \frac{20}{5} = 4 , \text{m/s}^2 ]

Using the acceleration found in the previous step in the first equation:

[ T = 3a = 3 \times 4 = 12 , \text{N} ]

For the 2 kg mass on the inclined plane:

• Weight acting downwards: (2g)
• Normal force acting perpendicular to the incline: (R)
• Frictional force acting up the incline: (\mu R), where (\mu = \frac{1}{2})

These forces should be illustrated in a free body diagram.

Using the equations of motion for the inclined plane:

The net force down the plane:

[ 2g \sin(\theta) - \mu R = 2a ]

With ( R = 2g \cos(\theta) ) and substituting ( \sin(\theta) = \frac{4}{5} ) and ( \cos(\theta) = \frac{3}{5} ):

[ 2g \sin(\theta) - \frac{1}{2}(2g \cos(\theta)) = 2a ]

Solving this will give:

[ a = 5 , \text{m/s}^2 ]