1. Ana wong sing ana ing njero lift sing obah munggah kanthi kecepatan tetep. The bobot wong kasebut ana ing 800 N. Tali lift langsung pedhot, mula lift kasebut tiba. Temtokake kekuatan normal tumindak saka lantai lift marang wong kasebut sadurunge lan sawise tali lift pedhot.
A. 800 N lan 0
B. 800 N lan 800 N
C. 1600 U lan 0
D. 1600 Lor lan 800 Lor
Dikenal:
Weight (w) = 800 Newton
Dikarepake: Gaya normal (N)
Solusi:
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 percepatan), 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 tiba gratis together, where the magnitude and the direction of their acceleration same as acceleration due to gravity. There is no normal force on the person.
Wangsulan sing bener yaiku 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 friction force experienced by the block.
A. 0.3 Lor
B. 1.4 N
C. 2.0 Lor
D. 3.6 Lor
Dikenal:
Massa (m) = 20 gram
Gaya (F) = 2 Newton
Dikarepake: Magnitude of friction force experienced by the block.
Solusi:
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 (Fdhuwit) same as the magnitude of the external force (F)
– The friction force (Fdhuwit) has opposite direction with the external force (F)
Apply Newton’s first law of motion :
∑F=0
F – Fdhuwit = 0
F = Fdhuwit
Fdhuwit = 2 Newton
Wangsulan sing bener yaiku 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 Lor
B. 300 N
C. 600 Lor
D. 900 Lor
Dikenal:
Weight of block (w) = 1350 Newton
hyp = 0.6 m
opp = 0.4 m
Dikarepake: The minimum force
Solusi:
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).
∑F=0
F – wx = 0
F = wx
If F = wx, banjur object start to moving upward at constant velocity.
wx = w sin θ = (1350)(2/3) = (2)(450) = 900 Newton
Wangsulan sing bener yaiku D.
4. Three forces, F1 = 22 N, F2 = 18 N lan F3 = 40 N act on a block. Which figure describes Newton’s first law.
Solusi:
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)
