1. Osoba je ve výtahu, který se pohybuje nahoru rychlostí konstantní rychlost, váha síla osoby je 800 N. Lano výtahu se okamžitě přetrhne, takže výtah spadne. Určete normálová síla jednalo se z patra výtahu na osobu těsně před a po přetržení lana výtahu.
A. 800 N a 0
B. 800 N a 800 N
cca 1600 N a 0
D. 1600 N a 800 N
Známý:
Weight (w) = 800 Newton
Hledám: Normálová síla (N)
Řešení:
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 zrychlení), so there is no net force to act on the person.
∑F = 0
S – š = 0
N = w
N = 800 Newtonů
After the elevator’s rope broke
After the elevator’s rope broke, the elevator and the person volný pád together, where the magnitude and the direction of their acceleration same as acceleration due to gravity. There is no normal force on the person.
Správná odpověď je 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 třecí síla experienced by the block.
A. 0.3 N
B. 1.4 N
C. 2.0 N
D. 3.6 N
Známý:
Hmotnost (m) = 20 gramů
Síla (F) = 2 Newtony
Hledám: Magnitude of friction force experienced by the block.
Řešení:
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 (Fpeníze) same as the magnitude of the external force (F)
– The friction force (Fpeníze) has opposite direction with the external force (F)
Apply Newton’s first law of motion :
∑F = 0
F – Fpeníze = 0
F=Fpeníze
Fpeníze = 2 Newtonů
Správná odpověď je 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
C. 600 N
D. 900 N
Známý:
Weight of block (w) = 1350 Newton
hyp = 0.6 m
opp = 0.4 m
Hledá se: The minimum force
Řešení:
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
P – tx = 0
F = wx
If F = wx, pak object start to moving upward at constant velocity.
wx = w sin θ = (1350)(2/3) = (2)(450) = 900 Newton
Správná odpověď je D.
4. Three forces, F1 = 22 N, F2 = 18 N a F3 = 40 N act on a block. Which figure describes Newton’s first law.
Řešení:
Newton’s first law : Net force (ΣF) = 0.
A. F.1 + F2 - F3 = 22 N + 18 N – 40 N = 40 N – 40 N = 0
B. F.2 + F3 - F1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)
C. F.2 + F3 - F1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)
D. F.1 + F3 - F2 = 22 N + 40 N – 18 N = 62 N – 18 N = 44 N (leftward)
