Photo AI
Two blocks of masses 220 kg and 75 kg lie stationary on a ROUGH horizontal surface - NSC Technical Sciences - Question 2 - 2023 - Paper 1
Question 2
Two blocks of masses 220 kg and 75 kg lie stationary on a ROUGH horizontal surface. The blocks are connected by a light inextensible string. The string makes an angl... show full transcript
Worked Solution & Example Answer:Two blocks of masses 220 kg and 75 kg lie stationary on a ROUGH horizontal surface - NSC Technical Sciences - Question 2 - 2023 - Paper 1
Step 1
2.1 Draw a labelled free-body diagram indicating ALL the forces acting on the 220 kg block while it is accelerating.
Answer
The free-body diagram of the 220 kg block includes the following forces:
- Weight (W) acting downwards:
- Normal force (N) acting upwards: Equal in magnitude and opposite to weight.
- Applied force (F_a = 165 N) acting horizontally to the left.
- Tension (T) acting at an angle of 30° upwards to the right.
- Frictional force (F_f = 35 N) acting horizontally opposite to the applied force.
Step 2
Step 3
2.3 Calculate the magnitude of the tension in the string.
Answer
To find the tension (T), apply Newton's Second Law to the 220 kg block:
- The net force in the horizontal direction: [ F_{net} = F_a - F_f - T_x ] where [ T_x = T \cos(30°) ]
- Convert forces to components: [ 165 ext{ N} - 35 ext{ N} - T \cos(30°) = 220 ext{ kg} imes 0.4 ext{ m/s}^2 ]
- Solving for T:
[ 130 - T \cos(30°) = 88 \Rightarrow T \cos(30°) = 42
T = \frac{42}{\cos(30°)} \approx 48.28 ext{ N} ]
Step 4
2.4 Calculate the coefficient of kinetic friction, μk, of the 75 kg block.
Answer
Using the frictional force formula:
- The net force on the 75 kg block is: [ F_{net} = ma = 75 ext{ kg} \times 0.4 ext{ m/s}^2 = 30 ext{ N} ]
- The frictional force is expressed as: [ F_f = μ_k N ] where [ N = mg = 75 ext{ kg} × 9.81 ext{ m/s}^2 = 735.75 ext{ N} ]
- From the forces acting on the block: [ 30 = μ_k \times 735.75 \Rightarrow μ_k = \frac{30}{735.75} \approx 0.0407 ]
Step 5
2.5 How will the magnitude of the frictional force of the 75 kg block be affected if the string that joined the blocks was horizontal? Write down INCREASES, DECREASES or REMAINS THE SAME.
Answer
The magnitude of the frictional force of the 75 kg block will REMAIN THE SAME if the string was horizontal since the normal force acting on the block does not change, affecting the frictional force, which is dependent on the normal force.
Step 6
2.6 Explain the answer to 2.5 above.
Answer
The frictional force is calculated using the formula: [ F_f = μ_k N ] Thus, whether the string is at an angle or horizontal, as long as the normal force remains constant (assuming weight conditions are unchanged), the frictional force remains unchanged. The angle of the string affects tensions and forces therein but does not influence the normal force affecting the 75 kg block.