Solved problems in Newton’s laws of motion – Newton’s second law of motion
1. A 1 kg object accelerated at a constant 5 m/s2. Estimate the net force needed to accelerate the object.
ज्ञात :
वस्तुमान (m) = १ किलोग्रॅम
प्रवेग (a) = 5 m/s2
पाहिजे : net force (∑F)
उपाय:
We use Newton’s second law to get the net force.
Έ�F = ma
Έ�F = (1 kg)(5 m/s2) = ८० किलोग्रॅम मीटर/सेकंद2 = ४० न्यूटन
2. वस्तुमान of an object = 1 kg, net force ∑F = 2 Newton. Determine the magnitude and direction of the object’s acceleration….
ज्ञात :
वस्तुमान (m) = १ किलोग्रॅम
Net force (∑F) = 2 Newton
पाहिजे : The magnitude and direction of the acceleration (a)
उपाय:
a = ∑F / m
अ = २० / १०
a = ४ मी/से2
The direction of the acceleration = the direction of the net force (∑F)
3. Object’s mass = 2 kg, F1 = १० न्यूटन, F2 = 3 Newton. The magnitude and direction of the acceleration is…
ज्ञात :
वस्तुमान (m) = १ किलोग्रॅम
F1 = ४० न्यूटन
F2 = ४० न्यूटन
पाहिजे आहे : The magnitude and direction of the acceleration (a)
उपाय:
net force :
Έ�एफ = एफ1 - एफ2 = ४८ – ६.४ = ४१.६ न्यूटन
त्वरणाचे परिमाण :
a = ∑F / m
अ = २० / १०
a = ४ मी/से2
Direction of the acceleration = direction of the net force = direction of F1
4. Object’s mass = 2 kg, F1 = १० न्यूटन, F2 = 1 Newton. The magnitude and direction of the acceleration is…
ज्ञात :
वस्तुमान (m) = १ किलोग्रॅम
F2 = ४० न्यूटन
F1 = ४० न्यूटन
F1x = एफ1 कारण 60o = (10)(0.5) = 5 न्यूटन
पाहिजे : The magnitude and direction of the acceleration (a)
उपाय:
निव्वळ बल :
Έ�एफ = एफ1x - एफ2 = ४८ – ६.४ = ४१.६ न्यूटन
त्वरणाचे परिमाण :
a = ∑F / m
अ = २० / १०
a = ४ मी/से2
Direction of the acceleration = direction of the net force = direction of F1x
5. एफ1 = १० न्यूटन, F2 = १ न्यूटन, मीटर1 = २ किलोग्रॅम, मीटर2 = 2 kg. The magnitude and direction of the acceleration is…
ज्ञात :
वस्तुमान २ (मी)1) = ४ किलो
वस्तुमान २ (मी)2) = ४ किलो
F1 = ४० न्यूटन
F2 = ४० न्यूटन
पाहिजे : The magnitude and direction of the acceleration (a)
उपाय:
The net force :
Έ�एफ = एफ1 - एफ2 = ४८ – ६.४ = ४१.६ न्यूटन
त्वरणाचे परिमाण :
a = ∑F / (m1 + मी2)
a = 9 / (1 + 2)
अ = २० / १०
a = ४ मी/से2
The direction of the acceleration = the direction of the net force = direction of F1
6.
A 40-kg block accelerated by a force of 200 N. Acceleration of the block is 3 m/s2. Determine the magnitude of friction force experienced by the block.
अ. २८.८ एन
बी. ३०.० एन
सी. ३९.६ एन
डी. ४८.० एन
ज्ञात :
वस्तुमान (m) = १ किलोग्रॅम
बल (F) = 200 N
त्वरण (a) = 3 m/s²2
पाहिजे: घर्षण बल (Fg)
उपाय:
समीकरण न्यूटनचा गतीचा दुसरा नियम
Έ�F = ma
Έ�F = net force, m = mass, a = acceleration
The direction of force F rightward, the direction of friction force leftward (the direction of friction force is opposite with the direction of object’s motion).
Choose rightward as positive and leftward as negative.
Έ�F = ma
एफ – एफg = मा
200 - एफg = (40)(3)
200 - एफg = 120
Fg = ४ – २
Fg = ४० न्यूटन
योग्य उत्तर D आहे.
7. Block A with a mass of 100-gram place above block B with a mass of 300 gram, and then block b pushed with a force of 5 N vertically upward. Determine the सामान्य बल exerted by block B on block A.
अ. २८.८ एन
बी. ३०.० एन
सी. ३९.६ एन
डी. ४८.० एन
ज्ञात :
बल (F) = १२ न्यूटन
ब्लॉक A चे वस्तुमान (मी)A) = ३०० ग्रॅम = ०.३ किलो
ब्लॉक B चे वस्तुमान (मी)B) = ३०० ग्रॅम = ०.३ किलो
गुरुत्वाकर्षणाचा प्रवेग (g) = 10 m/s2
वजन of block A (wA) = (०.३ किलोग्रॅम)(१० मीटर/सेकंद2) = ८० किलोग्रॅम मीटर/सेकंद2 = ४० न्यूटन
ब्लॉक B चे वजन (wB) = (०.३ किलोग्रॅम)(१० मीटर/सेकंद2) = ८० किलोग्रॅम मीटर/सेकंद2 = ४० न्यूटन
पाहिजे आहे : Normal force exerted by block B to block A
उपाय:
There are several forces that act on both block, as shown in figure.
F = push force (act on block B)
wA = weight of block A (act on block A)
wB = weight of block B (act on block B)
NA = normal force exerted by block B on block A (Act on block A)
NA’ = normal force exerted by block A on block B (Act on block B)
Apply Newton’s second law of motion on both blocks :
Έ�F = ma
एफ – डब्ल्यूA - डब्ल्यूB + एनA - एनA' = (मीA + मीB) a
NA आणि एनA’ are action-reaction forces that have the same magnitude but opposite in direction so eliminated from the equation.
एफ – डब्ल्यूA - डब्ल्यूB = (मीA + मीB) a
५ – १ – ३ = (०.१ + ०.३) अ
५ – ४ = (०.४) अ
४० = (५) अ
अ = २० / १०
a = ४ मी/से2
Apply Newton’s second law of motion on block A :
Έ�F = ma
NA - डब्ल्यूA = मीA a
NA – 1 = (0.1)(2.5)
NA - 1 = 0.25
NA = २८ + ३
NA = ४० न्यूटन
योग्य उत्तर B आहे.
8. An object with weight of 4 N supported by a cord and pulley. A force of 2 N acts on the block and one end of the cord pulled by a force of 9 N. Determine the net force acts on object X.
A. 3 N upward
B. 4 N downward
C. 9 N upward
D. 9 N downward
ज्ञात :
Weight of X (wX) = १० न्यूटन
Pull force (Fx) = १० न्यूटन
Tension force (FT) = १० न्यूटन
पाहिजे: Net force acts on object X
उपाय:
Vertically upward forces that act on object X :
The tension force has the same magnitude in all part of the cord. So the tension force is 9 N.
Vertically downward forces that act on object X :
There are two forces that act on object X and both forces are vertically downward, the horizontal component of weight wx and the horizontal component of force Fx.
Net force act on the object X :
FT - डब्ल्यूX - एफx = 9 – 4 – 2 = 9 – 6 = 3
The net force act on the object X is 3 Newton, vertically upward.
योग्य उत्तर A आहे.
9. An object initially at rest on a smooth horizontal surface. A force of 16 N acts on the object so the object accelerated at 2 m/s2. If the same object at rest on a rough horizontal surface so the friction force acts on the object is 2 N, then determine the acceleration of the object if the same force of 16 N acts on the object.
अ. १० मी/से2
बी. १२ मी/से2
सी. १५ मी/से2
डी. २० मी/से2
ज्ञात :
Force (F) = 16 Newton = 16 kg m/s2
त्वरण (a) = 2 m/s²2
घर्षण बल (F)पैसे) = 2 Newton = 2 kg m/s2
पाहिजे आहे : Object’s acceleration ?
उपाय:
Smooth horizontal surface (no friction force) :
Έ�F = ma
F = ma
16 = (m) 2
मी = ४० / १५६०
मी = 8 किलो
Mass of object is 8 kilogram.
Rough horizontal surface (there is a friction force) :
Έ�F = ma
एफ – एफपैसे = मा
४०० – १०० = ३०० अ
४८० = २४० अ
अ = २० / १०
a = ४ मी/से2
Object’s acceleration is 1.75 m/s2.
योग्य उत्तर A आहे.
10. Tom and Andrew push an object on the smooth floor. Tom push the object with a force of 5.70 N. If the mass of the object is 2.00 kg and acceleration experienced by the object is 2.00 ms-2, then determine the magnitude and direction of force act by Tom.
A. 1.70 N and its direction is opposite with force acted by Andre.w
B. 1.70 N and its direction same as force acted by Andrew
C. 2.30 N and its direction is opposite with force acted by Andrew.
D. 2.30 N and its direction same as force acted by Andrew.
ज्ञात :
Push force acted by Andrew (F1) = १० न्यूटन
वस्तूचे वस्तुमान (m) = १६ किलोग्रॅम
त्वरण (a) = 2.00 m/s²2
पाहिजे आहे : Magnitude and direction of force acted by Tom (F2)?
उपाय:
Apply Newton’s second law of motion :
Έ�F = ma
F1 + एफ2 = मा
5.70 + F2 = (2)(2)
5.70 + F2 = 4
F2 = ४ – २
F2 = – 1.7 Newton
Minus sign indicated that (F2) is opposite with push force act by Andrew (F1).
योग्य उत्तर A आहे.
11. If the mass of the block is the same, which figure shows the smallest acceleration?
उपाय
Net force A :
ΣF = 4 N + 2 N – 3 N = 6 N – 3 N = 3 Newton, leftward
Net force B :
ΣF = 2 N + 3 N – 4 N = 5 N – 4 N = 1 Newton, rightward
Net force C :
ΣF = 4 N + 3 N – 2 N = 7 N – 2 N = 5 Newton, rightward
Net force D :
ΣF = 3 N + 4 N + 2 N = 9 Newton, rightward
The equation of Newton’s second law :
ΣF = ma
a = ΣF / m
a = acceleration, ΣF = net force, m = mass
Based on the above formula, the acceleration (a) is directly proportional to the net force (ΣF) and inversely proportional to mass (m). If the mass of an object is the same, the greater the resultant force, the greater the acceleration or the smaller the resultant force, the smaller the acceleration.
Based on the above calculation, the smallest net force is 1 Newton so the acceleration is also smallest.
योग्य उत्तर B आहे.
12. Some forces act on an object with a mass of 20 kg, as shown in the figure below.
Determine the object’s acceleration.
ज्ञात :
वस्तूचे वस्तुमान (m) = १६ किलोग्रॅम
Net force (ΣF) = 25 N + 30 N – 15 N = 40 N
पाहिजे: Acceleration of an object
उपाय:
Object’s acceleration calculated using the equation of Newton’s second law :
ΣF = ma
a = ΣF / m = 40 N / 20 kg = 2 N/kg = 2 m/s2
13. Which statements below describes Newton’s third law?
(1) Passengers pushed forward when the bus braked suddenly
(2) Books on paper are not falling when the paper is pulled quickly
(3) When playing skateboard when the foot pushes the ground back then the skateboard will slide forward
(१) ओars pushed backward, boats moving forward
उपाय:
(2) Newton’s first law
(3) Newton’s third law
(4) Newton’s third law
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- वस्तुमान आणि वजन
- सामान्य शक्ती
- न्यूटनचा गतीचा दुसरा नियम
- घर्षण बल
- घर्षण बलाशिवाय क्षैतिज पृष्ठभागावरील गती
- घर्षण बलासह खडबडीत क्षैतिज पृष्ठभागावर समान त्वरणाने दोन वस्तूंची गती
- घर्षण बलाशिवाय उतरत्या पृष्ठभागावरील गती
- घर्षण बलासह खडबडीत कलत्या पृष्ठभागावरील गती
- लिफ्टमधील गती
- वस्तूंची गती दोऱ्या आणि कप्पीद्वारे जोडलेली असते.
- समान परिमाणाचे त्वरण असलेले दोन पिंड
- सपाट वळण घेणे – वर्तुळाकार गतीची गतिशीलता
- उतार असलेल्या वळणाला वळवणे – वर्तुळाकार गतीची गतिशीलता
- क्षैतिज वर्तुळात एकसमान गती
- एकसमान वर्तुळाकार गतीमधील अभिकेंद्री बल
