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2 years ago

A man lifts a 120 kg barbell 2 m above the ground . What is the gain in gravitational PE of the barbell?

Physics

1 answer:

SpyIntel [72]2 years ago

7 0

Answer:

2,352 Joules

Explanation:

At the ground, the barbell has a classical mechanical energy value of zero. There is no classical kinetic or potential energy for the barbell. The moment the man starts to lift the barbell, he does work on the barbell and transfers kinetic energy to it due to the motion. At its maximum height where the man lifts the barbell to a stop, the kinetic energy is zero because it transformed into gravitational potential energy stored in the gravitational field. Our reference point for potential was defined to be zero at the floor, therefore we can say that the gravitational potential energy at 2 meters is:

$U=mgh=(120kg)(9.8m/s^2)(2m)=2,352J$

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3 years ago

Phil is riding a scooter and pushes off the ground with his foot. this causes him to accelerate at 12 m /s. Phil weighs 600 N. h

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734.16 kg m/ $s^{2}$

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Acceleration: 12 m/ $s^{2}$

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Let x be the mass in newtons.

Let's convert: $\frac{1 kg}{9.80665 N}$ x $\frac{x}{600 N}$ = $\frac{600}{9.80665}$ = 61.18 kg

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3 years ago

Consider a semicircular ring of radius R. Its linear mass density varies as lambda =lambda not sin theta. Locate its centre of m

bearhunter [10]

Answer:

(0, πR/4)

Explanation:

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The mass of this short length is:

dm = λ ds

dm = (λ₀ sin θ) (R dθ)

dm = R λ₀ sin θ dθ

The x coordinate of the center of mass is:

X = ∫ x dm / ∫ dm

X = ∫₀ᵖ (R cos θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

X = R ∫₀ᵖ sin θ cos θ dθ / ∫₀ᵖ sin θ dθ

X = R ∫₀ᵖ ½ sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R ∫₀ᵖ 2 sin 2θ dθ / ∫₀ᵖ sin θ dθ

X = ¼R (-cos 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

X = ¼R (-cos 2π − (-cos 0)) / (-cos π − (-cos 0))

X = ¼R (-1 + 1) / (1 + 1)

X = 0

The y coordinate of the center of mass is:

Y = ∫ y dm / ∫ dm

Y = ∫₀ᵖ (R sin θ) (R λ₀ sin θ dθ) / ∫₀ᵖ R λ₀ sin θ dθ

Y = R ∫₀ᵖ sin² θ dθ / ∫₀ᵖ sin θ dθ

Y = R ∫₀ᵖ ½ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R ∫₀ᵖ (1 − cos 2θ) dθ / ∫₀ᵖ sin θ dθ

Y = ½R (θ − ½ sin 2θ)|₀ᵖ / (-cos θ)|₀ᵖ

Y = ½R [(π − ½ sin 2π) − (0 − ½ sin 0)] / (-cos π − (-cos 0))

Y = ½R (π − 0) / (1 + 1)

Y = ¼πR

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Explanation:

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3 years ago

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