1. Persuna tinsab f’lift li jiċċaqlaq ’il fuq b’ veloċità kostanti. il piż tal-persuna hija 800 N. Immedjatament il-ħabel tal-lift jinkiser, u għalhekk il-lift jaqa'. Iddetermina l- saħħa normali aġixxa mill-art tal-lift lill-persuna eżatt qabel u wara li nqata' l-ħabel tal-lift.
A. 800 N u 0
B. 800 N u 800 N
Ċ. 1600 N u 0
D. 1600 N u 800 N
Magħruf:
Weight (w) = 800 Newton
Mfittxija: The normal force (N)
Soluzzjoni:
Before the elevator’s rope broke
When the person stands on the floor of the elevator, weight acts on the person where the direction of the person is downward. That person at rest so that there must a normal force acts on the person, where the direction of the normal force is upward and the magnitude of the normal force same as the magnitude of the weight.
Because the person is at rest in the elevator and the elevator moves at a constant speed (no aċċelerazzjoni), so there is no net force to act on the person.
∑F = 0
N – w = 0
N = w
N = 800 Newton
After the elevator’s rope broke
After the elevator’s rope broke, the elevator and the person waqgħa ħielsa together, where the magnitude and the direction of their acceleration same as acceleration due to gravity. There is no normal force on the person.
It-tweġiba t-tajba hija A.
2. A block with a mass of 20 gram moves at a constant velocity on a rough horizontal floor at a constant velocity if there is an external force of 2 N acts on the block. Determine the magnitude of the forza ta 'frizzjoni experienced by the block.
A. 0.3 N
B. 1.4 N
Ċ 2.0 N
D. 3.6 Tramuntana
Magħruf:
Massa (m) = 20 gramma
Forza (F) = 2 Newton
Mfittxija: Magnitude of friction force experienced by the block.
Soluzzjoni:
Based on Newton’s first law of motion, if a block moves at a constant velocity, then the block has no acceleration. The block moves at a constant velocity, and there is no acceleration if :
– The magnitude of friction force (Ffric) same as the magnitude of the external force (F)
– The friction force (Ffric) has opposite direction with the external force (F)
Apply Newton’s first law of motion :
ΣF=0
F – Ffric = 0
F = Ffric
Ffric = 2 Newton
It-tweġiba t-tajba hija C.
3. A smooth inclined plane with the length of 0.6 m and height of 0.4 m. A block with the weight of, 1350 N will move upward using the inclined plane. Determine the magnitude of force need to move the block.
A. 100 N
B. 300 N
Ċ 600 N
D. 900 Tramuntana
Magħruf:
Weight of block (w) = 1350 Newton
hyp = 0.6 m
opp = 0.4 m
Meħtieġ: The minimum force
Soluzzjoni:
hyp = ac = 0.6 m
opp = bc = 0.4 m
Sin θ = bc / ac = 0.4 / 0.6 = 4/6 = 2/3
Based on Newton’s first law of motion, the block start to moves upward then the external force (F) minimal same as the horizontal component of weight (wx).
ΣF=0
F – wx = 0
F = wx
If F = wx, Imbagħad object start to moving upward at constant velocity.
wx = w sin θ = (1350)(2/3) = (2)(450) = 900 Newton
It-tweġiba t-tajba hija D.
4. Three forces, F1 = 22 N, F2 = 18-il Tramuntana u Tramuntana3 = 40 N act on a block. Which figure describes Newton’s first law.
Soluzzjoni:
Newton’s first law : Net force (ΣF) = 0.
A. F1 +F2 - F3 = 22 N + 18 N – 40 N = 40 N – 40 N = 0
B. F2 +F3 - F1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)
C. F2 +F3 - F1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)
D. F1 +F3 - F2 = 22 N + 40 N – 18 N = 62 N – 18 N = 44 N (leftward)
