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4. (a) The block P has a light pulley fixed to it - Leaving Cert Applied Maths - Question 4 - 2016
Question 4
4. (a) The block P has a light pulley fixed to it. The two blocks P and Q, of mass 40 kg and 30 kg respectively, are connected by a taut light inextensible string pa... show full transcript
Worked Solution & Example Answer:4. (a) The block P has a light pulley fixed to it - Leaving Cert Applied Maths - Question 4 - 2016
Step 1
Find (i) the acceleration of P and the acceleration of Q
Answer
To find the accelerations, we begin by applying Newton's second law to each mass.
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For block P:
- The forces acting on P along the incline are the tension (T) and the component of gravitational force parallel to the incline, which is (40g \sin(30°)).
- Applying Newton's second law:
[ 40g \sin(30°) - T - F_{friction} = 40a_P ]
where (F_{friction} = \frac{1}{4} \times 40g \cos(30°))
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For block Q:
- The forces acting on Q are the gravitational force and the tension in the string: [ 30g - T = 30a_Q ]
Since block P and Q are connected, their accelerations are related by (a_P = a_Q = a). From these equations, we set up the simultaneous equations to solve for T and a:
After substituting the values: [ a = 1.92 , ext{m/s}^2 ] Thus, the accelerations are: [ a_P = 1.92 , ext{m/s}^2 ] and [ a_Q = 3.84 , ext{m/s}^2 ]
Step 2
Find (ii) the speed of P when it has moved 30 cm
Answer
To find the speed of P after moving 30 cm, we can use the kinematic equation: [ v^2 = u^2 + 2as ] Here, (u = 0) (as it starts from rest), (s = 0.3 , ext{m}), and (a = 1.92 , ext{m/s}^2):
Substituting the known values: [ v^2 = 0 + 2(1.92)(0.3) ] [ v^2 = 1.152 ] [ v = \sqrt{1.152} \approx 1.07 , ext{m/s} ]
Step 3
Find (i) the tension in the string
Answer
To calculate the tension in the string, we use the equations derived earlier:
Using the dynamics of both masses:
- For the 5 kg block: [ T - 5g = 5a ]
- For the 7 kg block: [ T - 7g = -7a , (deceleration) ]
Adding these, we can solve for T: [ 14g = 14 \left( \frac{a+b}{2} \right) ] [ T = \frac{70g}{11} \approx 62.36 , ext{N} ]
Step 4
Find (ii) the time taken until the moveable pulley reverses direction
Answer
Considering the initial upward velocity (u) of the moveable pulley is 0.8 m/s: Using the first equation of motion: [ v = u + at ] Setting v = 0 (velocity becomes zero at the topmost point) and a as negative: [ 0 = 0.8 - 0.89t ] [ t = \frac{0.8}{0.89} \approx 0.90 , ext{s} ]