牛頓運動定律—問題與解答

牛頓運動定律—問題與解答

Newton’s first law of motion

1. If no net force acts on an object, then :

(1) the object is not accelerated

(2)靜止物體

(3) the change of velocity of an object = 0

(4) the object can not travels at a 等速

Which statement is correct.

解決方案

The correct statement :

(1) The object is not accelerated

The net force causes acceleration of an object. So if no net force then objects is not accelerated.

(2) Object at rest

Newton’s first law of motion states that if no net force acts on an object then an object always at rest or the object is always traveling at a constant velocity.

(3) The change of velocity of an object = 0

Change of velocity = acceleration. No change of velocity means no acceleration. If no acceleration then no net force acts on an object.

The object in an elevator

2. The weight of a person in an elevator at rest = 500 N. 重力加速度 是 10 公尺/秒2. When lift accelerated, the tension force is 750 N. What is the acceleration of lift.

已知:

人的 重量 (w) = 500 Newton = 500 kg m s - 2 (lift at rest)

Acceleration of gravity (g) = 10 m s - 2

人的 質量 (m) = 500 / 10 = 50 kg

Tension force (T) = 750 N (lift accelerated)

Elevator’s mass ignored.

通緝: Acceleration of elevator

解決方案:

Elevator at rest, no acceleration (a = 0). Force acts upward has plus sign and force acts downward has minus sign.

ΣF = ma

T – w = 0

T = w

T = 500 牛頓

If the elevator accelerated downward then the tension force smallest then 500 N. Otherwise, if the elevator accelerated upward then the tension force larger then 500 N.

The tension force = 750 N because the elevator accelerated upward. Force acts upward has plus sign and force acts downward has minus sign.

T – w = ma

750 – 500 = 50a

250 = 50 a

一個 = 250 / 50

a = 5.0 m s - 2

3. An 60-kg person in an elevator accelerated downward at 3 m/s2. If acceleration due to gravity is 10 m/s2, what is the normal force exerted by elevator’s floor on person.

已知:

質量 (m) = 60 kg

Acceleration of person and elevator (a) = 3 m/s2

重力加速度 (g) = 10 公尺/秒²2

重量 (w) = mg = (60)(10) = 600 牛頓

通緝: 法向力 (N)

解決方案:

There are two forces acts on the person in the elevator, that is weight (w) of person and the normal force (N) exerted by the floor on the person. There are three vector quantities, that is weight (w), normal force (N) and acceleration of elevator, where weight acts downward, the normal force acts upward, acceleration of elevator is downward. Vector quantities that act downward have plus sign and vector 數量 that act upward have minus sign.

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∑F = ma

w – N = (60)(3)

600 – N = 180

N = 600 – 180

N = 420 牛頓

4. A 40-kg object in an elevator accelerated upward. If the elevator’s floor exerts 520 N on object and acceleration due to gravity is 10 m/s2. What is the acceleration of the elevator?

已知:

質量 (m) = 40 kg

Normal force (N) = 520 N

重力加速度 (g) = 10 公尺/秒²2

weight (w) = m g = (40)(10) = 400 N

招募: Acceleration of elevator

解決方案:

∑F = ma

400 – 520 = (40)(a)

-120 = (40)(a)

a = -120/40

a = -3 公尺/秒2

Acceleration of elevator is 3 m/s2. Minus sign indicates that elevator travels upward.

5. An 60-kg object in an elevator accelerated downward at 3 m/s2. What is the force exerted by object on the elevator’s floor.

已知:

質量 (m) = 60 kg

重量 (w) = mg = (60 kg)(10 m/s)2) = 600 kg m/s2 = 600 牛頓

Acceleration of elevator (a) = 3 m/s2, downward

招募: Force exerted by object on the elevator’s floor.

解決方案:

Elevator accelerated downward at 3 m/s. Force acts downward has plus sign and force acts upward has minus sign.

w – N = ma

N = w – ma

N = 600 – (60)(3)

N = 600 – 180

N = 420 牛頓

Force exerted by the object on the elevator’s floor = 420 N.

6. Two blocks are connected by a cord running over a pulley. Ignore the mass of the cord and pulley and any friction in the pulley. Mass of block A is 6 kg and mass of block B is 2 kg. Acceleration due to gravity is 10 m/s2. What is the tension force?

已知:Newton's laws of motion – problems and solutions 1

mA = 6 kg,mB = 2 kg,g = 10 m/s2

wA =米A g = (6 kg)(10 m/s2) = 60 kg m/s2

wB =米B g = (2 kg)(10 m/s2) = 20 kg m/s2

招募: tension force (T) ?

解決方案:

wA > wB 以便 mA 向下移動,MB moves upward.

牛頓第二定律:

ΣF = ma

wA -wB = (米A + 米B)的

60 – 20 = (6 + 2) a

40 = (8) a

a = 40 / 8 = 5 公尺/秒2

Tension force :

mA 向下移動:

wA - T.A =米A a

60 - 噸A = (6)(5)

60 - 噸A = 30

TA = 60 - 30

T2 = 30 牛頓

mB 向上移動:

TB -wB =米B a

TB – 20 = (2)(5)

TB - 20 = 10

TB = 10 + 20

T1 = 30 牛頓

Tension force (T) = 30 Newton.

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7. Mass of object A = 5 kg, acceleration due to gravity (g) = 10 m s-2. Object A moves downward at 2.5 m.s-2. What is the mass of B ?

已知:Newton's laws of motion – problems and solutions 2

質量 A (mA) = 5 kg

重力加速度 (g) = 10 公尺/秒²2

Acceleration of object A (a) = 2.5 m/s2

Weight A (wA) = (mA)(g) = (5)(10) = 50 Newton

招募: 物體 B 的質量 (m)B)

解決方案:

Block A moves downward so weight of object A (wA) larger than weight of object B (wB).

應用牛頓第二定律:

ΣF = ma

wA -wB = (米A + 米B)的

50 – (mB)(10) = (5 + mB)(2.5)

50-10 mB = 12.5 + 2.5 mB

50 – 12.5 = 2.5 mB + 10 米B

37.5 = 12.5 米B

mB = 3公斤

8. Acceleration due to gravity is 10 m/s2. What is the tension force.

已知:

物體 1 的質量 (m)1) = 2 kgNewton's laws of motion – problems and solutions 3

物體 2 的質量 (m)2) = 3 kg

重力加速度 (g) = 10 公尺/秒²2

Weight 1 (w1) = (m1)(g) = (2 kg)(10 m/s2) = 20 kg m/s2

Weight 2 (w2) = (m2)(g) = (3 kg)(10 m/s2) = 30 kg m/s2

招募: tension force (T)

解決方案:

w2 > w1 以便 m2 moves downward and m1 moves upward.

Newton’s second law of motion :

ΣF = ma

w2 -w1 = (米1 + 米2)的

30 – 20 = (2 + 3)a

10 = (5) a

a = 10 / 5 = 2 公尺/秒2.

Tension force ?

m2 向下移動

w2 - T.2 =米2 a

30 - 噸2 = (3)(2)

30 - 噸2 = 6

T2 = 30 - 6

T2 = 24 牛頓

m1 向上移動

T1 -w1 =米1 a

T1 – 20 = (2)(2)

T1 - 20 = 4

T1 = 20 + 4

T1 = 24 牛頓

Tension force (T) = 24 Newton.

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  4. 問題: How does Newton’s third law explain the action of a rocket in space? 答: Newton’s third law states that for every action, there’s an equal and opposite reaction. A rocket in space propels itself by expelling exhaust gases backward. The action is the gas moving backward, and the reaction is the rocket moving forward.
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