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A block, of mass 8 kg, is placed on a rough horizontal surface - NSC Physical Sciences - Question 2 - 2018 - Paper 1

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A block, of mass 8 kg, is placed on a rough horizontal surface. The 8 kg block, which is connected to a 2 kg block by means of a light inextensible string passing ov... show full transcript

Worked Solution & Example Answer:A block, of mass 8 kg, is placed on a rough horizontal surface - NSC Physical Sciences - Question 2 - 2018 - Paper 1

Step 1

State Newton's Second Law in words.

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Answer

Newton's Second Law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In other words, the greater the force applied to an object, the greater its acceleration, while a larger mass will result in a smaller acceleration for the same amount of force.

Step 2

Draw a labelled free-body diagram for the 8 kg block.

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Answer

The free-body diagram for the 8 kg block should include:

  1. The weight of the block acting downwards (W = mg, where m = 8 kg and g = 9.81 m/s²).
  2. The normal force acting upwards.
  3. The tension in the string acting at an angle of 15° above the horizontal.
  4. The frictional force acting opposite to the direction of motion.

Step 3

Give a reason why the system is NOT in equilibrium.

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Answer

The system is not in equilibrium because the 8 kg block is experiencing a net acceleration of 1.32 m/s². This acceleration indicates that there is a resultant force acting on the block, and thus the forces are unbalanced.

Step 4

Use the 2 kg mass to calculate the tension in the string.

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Answer

To calculate the tension in the string, we can analyze the forces acting on the 2 kg block:

  1. The net force acting on the block is given by: Fnet=maFnet=2kg×1.32m/s2=2.64NF_{net} = ma \rightarrow F_{net} = 2 kg \times 1.32 m/s² = 2.64 N

  2. The tension in the string can also be found using: T=mgFnetT=(2kg×9.81m/s2)2.64N=19.62N2.64N=16.98NT = mg - F_{net} \rightarrow T = (2 kg \times 9.81 m/s²) - 2.64 N = 19.62 N - 2.64 N = 16.98 N

Step 5

Calculate the kinetic frictional force between the 8 kg block and the horizontal surface.

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Answer

We calculate the kinetic frictional force ( f_k = ext{Friction Coefficient} \times N

To find the normal force (N), we have: N=mg=8kg×9.81m/s2=78.48NN = mg = 8 kg \times 9.81 m/s² = 78.48 N

If we assume the coefficient of kinetic friction is known (( , \mu_k )), then:

fk=μk×N=μk×78.48Nf_k = \mu_k \times N = \mu_k \times 78.48 N

Based on the question’s data, if ( , \mu_k ) is calculated from earlier results, use that specific value, or denote the method to find it.

Step 6

Give a reason for this statement.

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Answer

The kinetic frictional force changes as the 8 kg block moves from B to C because the nature of the surfaces in contact may change. Different materials have varying coefficients of friction, which will result in a change in the frictional force experienced by the block as it transitions from one surface to another.

Step 7

Will the kinetic frictional force, calculated in QUESTION 2.3.3 above, change? Choose from: YES or NO. Give a reason for the answer.

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Answer

YES. The kinetic frictional force will change because it depends on the properties of the surfaces in contact. If the horizontal surface the 8 kg block is moving on is changed to a different material, the coefficient of kinetic friction will likely differ, resulting in a different kinetic frictional force.

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