1. Mmadụ nọ n'ime igwe mbuli elu nke na-arị elu n'ebe a ọsọ mgbe niile. The ibu ibu nke onye ahụ bụ 800 N. Ozugbo eriri ebuli ahụ gbajiri, ebuli ahụ ga-ada. Chọpụta ihe kpatara ya. ike nkịtị site n'ala ụlọ elu ahụ mee ihe nye onye ahụ tupu na mgbe eriri ụlọ elu ahụ gbawara.
A. 800 N na 0
B. 800 N na 800 N
C. 1600 N na 0
D. 1600 N na 800 N
A maara:
Weight (w) = 800 Newton
Chọrọ: The normal force (N)
Ngwọta:
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 ọganihu), 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 free-ada together, where the magnitude and the direction of their acceleration same as acceleration due to gravity. There is no normal force on the person.
Azịza ya bụ 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 ike esemokwu experienced by the block.
A. 0.3 N
B. 1.4 N
C. 2.0 N
D. 3.6 N
A maara:
Ibu (m) = gram 20
Ike (F) = 2 Newton
Chọrọ: Magnitude of friction force experienced by the block.
Ngwọta:
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 (Ffrik) same as the magnitude of the external force (F)
– The friction force (Ffrik) has opposite direction with the external force (F)
Apply Newton’s first law of motion :
ΣF=0
F – Ffrik = 0
F = Ffrik
Ffrik = 2 Newton
Azịza ya ziri ezi bụ 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
A maara:
Weight of block (w) = 1350 Newton
hyp = 0.6 m
opp = 0.4 m
A chọrọ: The minimum force
Ngwọta:
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, mgbe ahụ, object start to moving upward at constant velocity.
wx = w sin θ = (1350)(2/3) = (2)(450) = 900 Newton
Azịza ya ziri ezi bụ D.
4. Three forces, F1 = 22 N, F2 = 18 N na F3 = 40 N act on a block. Which figure describes Newton’s first law.
Ngwọta:
Newton’s first law : Net force (ΣF) = 0.
A. F1 +F2 - F.3 = 22 N + 18 N – 40 N = 40 N – 40 N = 0
B. F2 +F3 - F.1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)
C. F2 +F3 - F.1 = 18 N + 40 N – 22 N = 58 N – 22 N = 36 N (rightward)
D.F1 +F3 - F.2 = 22 N + 40 N – 18 N = 62 N – 18 N = 44 N (leftward)
