Motion on the inclined plane without the friction force – application of Newton’s law of motion problems and solutions

1. Box’s ibi- = 2 kg, isare nitori walẹ = 9.8m/s2. Find (a) the net force which accelerates the box downward (b) magnitude of the box’s isare.

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 1

ojutu

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 2

A mọ̀:

Ìwọ̀n (m) = 2 kg

Ìyára tí ó jẹ́ nítorí òòfà ilẹ̀ (g) = 9.8 m/s2

àdánù (w) = mg = (2)(9.8) = 19.6 Newtons

wx = w sin 30 = (19.6)(0.5) = 9.8 Newton

wy = w cos 30 = (19.6)(0.5√3) = 9.8√3 Newton

Ojutu:

(A) awọn net force which accelerates the box

Inclined plane is smooth, so there is no friction force. The only force which acts on the object is wx.

ΣF = wx

ΣF = 9.8 Newton

(B) magnitude of the acceleration

ΣF = ma

9.8 = (2) a

a = 9.8/2

a = 4.9 m/s2

Ìwọ̀n ìfàsẹ́yìn náà jẹ́ 4.9 m/s2, direction of the acceleration is downward.

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2. ofurufu ti idagẹrẹ is smooth so there is no ija edekoyede. Object’s mass is 3 kg, acceleration due to gravity is 9.8 m/s2. Determine the magnitude of the force F if (a) object is at rest (b) object is moving downward with constant acceleration 2 m/s2 (c) object is moving upward with a constant acceleration of 2 m/s2.

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 3

ojutu

Motion on incline plane without friction force - application of Newton's law of motion problems and solutions 4

A mọ̀:

Ìwọ̀n (m) = 3 kg

Ìyára tí ó jẹ́ nítorí òòfà ilẹ̀ (g) = 9.8 m/s2

Weight (w) = m g = (3)(9.8) = 29.4 Newton

wx = w sin 30 = (29.4)(0.5) = 14.7 Newton

wy = w cos 30 = (29.4)(0.5√3) = 14.7√3 Newton

Ojutu:

(a) The magnitude of the force F if an object is at rest

Newton ká akọkọ ofin of motion states that if an object is at rest, the net force acts on the object is zero.

ΣF=0

F – wx = 0

F = wx

F = 14.7 Newton

(b) The magnitude of the force F if an object is moving downward at a constant 2 m/s2

ΣF = ma

wx – F = m a

14.7 – F = (3)(2)

14.7 – F = 6

F = 14.7– 6

F = 8.7 Newton

(c) The magnitude of the force F if an object is moving upward at a constant 2 m/s2

ΣF = ma

F – wx = ma

F – 14.7 = (3)(2)

F – 14.7 = 6

F = 14.7 + 6

F = 20.7 Newton

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  1. Ibi ati iwuwo
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  3. Newton ká keji ofin ti išipopada
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  8. Ìṣípò lórí ọkọ̀ òfurufú onítẹ̀sí pẹ̀lú agbára ìfọ́kànsí
  9. Ìṣípò nínú ategun kan
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  11. Ara meji pẹlu iwọn iyara kanna
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