A force F, with a magnitude of 193,19 N, is applied at an angle of 20° with the horizontal to a block of mass 55 kg. The block moves at a CONSTANT velocity over a rough horizontal surface, PQ, as shown in the diagram below. The coefficient of kinetic friction between the block and the surface is 0,3. 2.1 Draw a labelled free-body diagram showing ALL the forces acting on the block. 2.2 State Newton's First Law of Motion in words. 2.3 Write down the magnitude of the net horizontal force acting on the block. Use a physics equation to explain the answer. 2.4 Calculate the magnitude of: 2.4.1 Vertical component of F 2.4.2 Normal force 2.4.3 Frictional force 2.5 When the same force F, is applied at an angle of 0° to the horizontal over the same rough horizontal surface, the magnitude of the normal force decreases. Explain why the magnitude of the normal force decreases.

NSC Technical Sciences Different kinds of forces: A force F, with a magnitude of 1

A force F, with a magnitude of 193,19 N, is applied at an angle of 20° with the horizontal to a block of mass 55 kg - NSC Technical Sciences - Question 2 - 2024 - Paper 1

Question 2

A force F, with a magnitude of 193,19 N, is applied at an angle of 20° with the horizontal to a block of mass 55 kg. The block moves at a CONSTANT velocity over a ro... show full transcript

Worked Solution & Example Answer:A force F, with a magnitude of 193,19 N, is applied at an angle of 20° with the horizontal to a block of mass 55 kg - NSC Technical Sciences - Question 2 - 2024 - Paper 1

Step 1

2.1 Draw a labelled free-body diagram showing ALL the forces acting on the block.

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Answer

In the free-body diagram, the following forces should be depicted:

Weight of the block ( $F_g = mg = 55 ext{ kg} imes 9.81 ext{ m/s}^2 = 539.55 ext{ N}$ ) acting downward.
Normal force (N) acting upward.
Applied force (F) at an angle of 20° to the horizontal, with its components:
- Vertical component ( $F_y$ ) = $F imes ext{sin}(20°)$
- Horizontal component ( $F_x$ ) = $F imes ext{cos}(20°)$
Frictional force (f) acting opposite to the direction of motion, calculated as:
- $f_k = ext{friction coefficient} imes N$

Step 2

2.2 State Newton's First Law of Motion in words.

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Answer

Newton's First Law of Motion states that an object continues in a state of rest or uniform motion (moving with a constant velocity) unless it is acted upon by a net external force.

Step 3

2.3 Write down the magnitude of the net horizontal force acting on the block. Use a physics equation to explain the answer.

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Since the block is moving at a constant velocity, the net horizontal force ( $F_{ ext{net}}$ ) acting on it is zero: $F_{ ext{net}} = F_x - f_k = 0 \\ \Rightarrow F_x = f_k$ This means the applied force in the horizontal direction equals the frictional force.

Step 4

2.4 Calculate the magnitude of:

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Step 5

2.4.1 Vertical component of F

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The vertical component ( $F_y$ ) is calculated as follows: $F_y = F imes ext{sin}(20°) = 193.19 ext{ N} imes ext{sin}(20°) = 66.07 ext{ N}$

Step 6

2.4.2 Normal force

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The normal force ( $N$ ) can be found using the equilibrium of vertical forces: $N = F_g - F_y = 539.55 ext{ N} - 66.07 ext{ N} = 473.48 ext{ N}$

Step 7

2.4.3 Frictional force

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The frictional force can be calculated using the normal force: $f_k = ext{friction coefficient} imes N = 0.3 imes 473.48 ext{ N} = 142.04 ext{ N}$

Step 8

2.5 Explain why the magnitude of the normal force decreases.

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When the force F is applied at an angle of 0°, all of its weight contributes to the vertical forces without any vertical component acting upwards. This means that the entire weight of the block contributes to the normal force, increasing its magnitude compared to the scenario where it is applied at an angle, which reduces the effective weight as the vertical component of the applied force counteracts the weight.

A force F, with a magnitude of 193,19 N, is applied at an angle of 20° with the horizontal to a block of mass 55 kg - NSC Technical Sciences - Question 2 - 2024 - Paper 1

Worked Solution & Example Answer:A force F, with a magnitude of 193,19 N, is applied at an angle of 20° with the horizontal to a block of mass 55 kg - NSC Technical Sciences - Question 2 - 2024 - Paper 1

2.1 Draw a labelled free-body diagram showing ALL the forces acting on the block.

2.2 State Newton's First Law of Motion in words.

2.3 Write down the magnitude of the net horizontal force acting on the block. Use a physics equation to explain the answer.

2.4 Calculate the magnitude of:

2.4.1 Vertical component of F

2.4.2 Normal force

2.4.3 Frictional force

2.5 Explain why the magnitude of the normal force decreases.

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