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
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 will be represented:
Weight (W) acting downwards:
W=mg=55extkgimes9.81extm/s2=539.55extN
Normal force (N) acting upwards:
Applied force (F) at an angle of 20° to the horizontal, which can be broken into two components:
Horizontal component (Fx): Fx=Fimesextcos(20°)
Vertical component (Fy): Fy=Fimesextsin(20°)
Frictional force (f) opposing the motion, acting horizontally:
f = extμimesN where μ = 0.3 is the coefficient of friction.
Step 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 (with a constant velocity) unless it is acted upon by a net external or unbalanced force. In simpler terms, if there is no net force acting on an object, it will not change its state of motion.
Step 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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Answer
Since the block is moving at a constant velocity, the net horizontal force acting on it must be zero. This can be expressed as:
Fextnet=Fextx−f=0
Thus, the applied horizontal force must be equal to the frictional force that opposes it.
Step 4
Calculate the magnitude of:
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Step 5
Explain why the magnitude of the normal force decreases.
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Answer
When the force F is applied at an angle of 0°, the entire force is applied horizontally. The vertical component (which contributes to the normal force) becomes zero. Thus, there is no additional upward force contributing to the normal force, which reduces its magnitude compared to when the force is applied at an angle.