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Two blocks, A, of mass 4 kg, and B, of mass 9 kg, are connected by a light inextensible string - NSC Physical Sciences - Question 2 - 2023 - Paper 1

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Two blocks, A, of mass 4 kg, and B, of mass 9 kg, are connected by a light inextensible string. The blocks are held at rest on a plane which is inclined at an angle ... show full transcript

Worked Solution & Example Answer:Two blocks, A, of mass 4 kg, and B, of mass 9 kg, are connected by a light inextensible string - NSC Physical Sciences - Question 2 - 2023 - Paper 1

Step 1

2.1 State Newton's Second Law of Motion in words.

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Answer

Newton's Second Law of Motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. This can be expressed mathematically as:

Fnet=mimesaF_{net} = m imes a

where FnetF_{net} is the net force, mm is the mass, and aa is the acceleration.

Step 2

2.2 Draw a labelled free-body diagram showing all the forces acting on block A.

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Answer

In the free-body diagram for block A, the following forces should be labeled:

  • Normal force (FNF_N) acting perpendicular to the inclined plane.
  • Gravitational force (WW) acting downwards, which can be resolved into two components: one parallel to the incline (Wimesextsin(35°)W imes ext{sin}(35°)) and the other perpendicular (Wimesextcos(35°)W imes ext{cos}(35°)).
  • Tension in the string (TT) acting upwards along the incline.
  • Kinetic frictional force (fkf_k) acting downwards along the incline, opposing the motion.

Step 3

2.3.1 The tension in the string.

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Answer

To find the tension in the string (T) for block A, we start with the equation:

Fnet=TfkWimesextsin(35°)=mimesaF_{net} = T - f_k - W imes ext{sin}(35°) = m imes a

Substituting the known values, we have: T5,884gimesextsin(35°)=4imes2T - 5,88 - 4g imes ext{sin}(35°) = 4 imes 2

Calculating the components gives: T=4imes2+5,88+4imes9.81imesextsin(35°)T = 4 imes 2 + 5,88 + 4 imes 9.81 imes ext{sin}(35°) T=8+5,88+4imes9.81imes0,5736T = 8 + 5,88 + 4 imes 9.81 imes 0,5736 T36,36extNT ≈ 36,36 ext{ N}

Step 4

2.3.2 Force F.

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Answer

Using the equation for block B:

FTfB=mBimesaF - T - f_B = m_B imes a

where fBf_B is the kinetic friction for block B. Now substituting:

F36,3613,23=9imes2F - 36,36 - 13,23 = 9 imes 2

Solving for F gives: F=18+36,36+13,23F = 18 + 36,36 + 13,23 F118,18extNF ≈ 118,18 ext{ N}

Step 5

2.4.1 How will this change the kinetic frictional force acting on block A?

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Answer

The kinetic frictional force acting on block A will DECREASE as the angle of inclination is decreased.

Step 6

2.4.2 Explain the answer to QUESTION 2.4.1.

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Answer

As the angle of the incline decreases, the normal force acting on block A also decreases. Since the kinetic frictional force is directly proportional to the normal force, this means the frictional force will decrease. The formula for kinetic friction is:

fk=extμkimesFNf_k = ext{μ}_k imes F_N

where FNF_N is the normal force, which is reduced with a lower incline angle.

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