Application of conservation of mechanical energy for projectile motion – problems and solutions

1. یو لات شوی فوټبال په ځمکه کې په داسې زاویه کې پریږدي چې θ = 30 ويo with the initial velocity of 10 m/s. Ball’s ډله = 0.1 کیلوګرامه سرعت د جاذبې له امله 1 دی0 م / س2. Determine (a) The د جاذبې احتمالي انرژي at the highest point (b) The highest point or the maximum height

پیژندل شوی:

وزن (متر) = 0.1 کیلوګرامه

The initial velocity (vo) = ۲ متره/ ثانیې

Angle = 30o

د جاذبې قوې له امله سرعت (g) = 10 m/s2

حل:

(a) The gravitational potential energy

Application of conservation of mechanical energy for projectile motion – problems and solutions 1

Calculate the horizontal component (vox) and the vertical component (voy) of initial velocity.

Application of conservation of mechanical energy for projectile motion – problems and solutions 2Application of conservation of mechanical energy for projectile motion – problems and solutions 2vox = ویo کاس θ = (۲۰)(د ۳۰ په اندازهo) = (10)(0.5√3) = 5√3 m/s

voy = ویo د ګناه θ = (۱۰)(ګناه ۶۰o) = (10)(0.5) = 5 m/s

لومړنۍ میخانیکي انرژي

د لومړنیو میخانیکي انرژي (MEo) = متحرکه انرژي (KE)

MEo = KE = ½ m vo2 = ½ (۱)(۰.۱)2 = ½ (0.1)(100) = ½ (10) = 5 Joule

The final mechanical energy

Kinetic energy at the highest point :

KE = ½ m vox2 = ½ (0.1)(5√3)2 = ½ (0.1)((25)(3)) = ½ (0.1)(75) = 3.75 Joule

Principle of conservation of mechanical energy

The initial mechanical energy (MEo) = the final mechanical energy (MEt)

KE = PE + KE

5 = EP + 3.75

PE = 5 – 3.75 = 1.25 Joule

The gravitational potential energy at the highest point is 1.25 Joule.

(b) The highest point or the maximum height

PE = m g h

1.25 = (0.1)(10) h

1.25 = h

The maximum height is 1.25 meters.

هم وګوره  د ثقل مرکز - ستونزې او حل لارې

2. A 0.1-kg ball projected horizontally with initial velocity vo = 10 m/s from a building 10 meter high. Acceleration due to gravity is 10 m/s2. Determine ball’s kinetic energy when it hits the ground.

پیژندل شوی:

وزن (متر) = 0.1 کیلوګرامه

لومړنی سرعت (vo) = ۲ متره/ ثانیې

د جاذبې قوې له امله سرعت (g) = 10 m/s2

The change in height (h) = 10 – 2 = 8 m

SE busca: kinetic energy at 2 meters above the ground

حل:

The gravitational potential energy (PE) = m g h = (0.1)(10)(10) = 10 Joule

The initial kinetic energy (KE)= ½ m vo2 = ½ (۱)(۰.۱)2 = ½ (0.1)(100) = ½ (10) = 5 Joule

The final kinetic energy = the initial gravitational potential energy + the initial kinetic energy = 10 + 5 = 15 Joule

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