Masses of 5 kg and 3 kg are connected by a light inelastic string which passes over a smooth light pulley - Leaving Cert Applied Maths - Question 4 - 2021

From the second part of Newton's second law, we can establish the equations:

1. For the 5 kg mass:
   ( 5g - T = 5a )
2. For the 3 kg mass:
   ( T - 3g = 3a )

We can solve these two equations: Replacing ( T ) from the second equation: [ T = 3g + 3a ]
Substituting into the first equation: [ 5g - (3g + 3a) = 5a ]
[ 2g = 8a ]
This gives us the acceleration ( a = \frac{g}{4} ), and subsequently, substituting back, ( T = 3g + 3 \left(\frac{g}{4}\right) = 37.5 \text{ N} ).

From our previous calculations, we established that: [ a = 2.5 , \text{m/s}^2 ]
This indicates the common acceleration of the masses in the system.

We can use the kinematic equation: [ v^2 = u^2 + 2as ]
Where ( u = 0 ), ( a = 2.5 \text{ m/s}^2 ), and ( s = 2 ext{ m} ). Substituting the values, we get: [ v^2 = 0 + 2 \times 2.5 \times 2 ]
[ v^2 = 10 ]
Thus, ( v = \sqrt{10} \approx 3.16 , \text{m/s} ).

For the 5.2 kg mass on the incline:

• Normal force ( R )
• Frictional force ( \mu R )
• Tension ( T ) acting upwards
• Weight component down the slope: ( 5.2g \sin \theta )

For the 4 kg mass:

• Weight ( 40 ext{ N} ) acting downwards
• Tension ( T ) acting upwards.

Using Newton's second law for the 4 kg mass: [ 40 - T = 4a ]
And for the 5.2 kg mass: [ T = 52 \sin \theta - \mu R = 5.2a ] Utilizing the given ( \tan \theta = \frac{5}{12} ) to obtain the necessary values and plugging them in: Solving for ( a ) results in: [ a = \frac{20}{23} , \text{m/s}^2 \approx 0.87 , \text{m/s}^2 ].