ضغط جسم كتلته 1 كغم على نابض ثابت مرونته 10 نيوتن مثبت على حائط لمسافة 10سم ثم تحرك لينزلق على سطح افقي املس حتى وصل الى منحدر خشن يميل بزاوية 37 مع الاسف ومعامل احتكاك سطحه 0,5 فاذا علمت ان الجسم وصل اسفل السطح المائل بسرعة 11 متر / ثانية احسب ارتفاع السطح

The energy stored in the spring = the kinetic energy at the bottom of the incline

1/2 kx² = 1/2 mv²

kx² = mv²

(500 N/m) (0.75 m)² = (10 kg) v²

v ≈ 5.3 m/s

The energy stored in the spring = the initial potential energy - work done by friction

1/2 kx² = mgh - W

1/2 (500 N/m) (0.75 m)² = (10 kg) (9.8 m/s²) (2.0 m) - W

W ≈ 55 J

Since the horizontal surface is frictionless, the object will have the same speed at the bottom of the incline as before.

v ≈ 5.3 m/s

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures

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