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A rope exerts a force F on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s2 near the surface of the earth. What is the kinetic energy (in J) of the crate when it is 4.0 m above the floor?

Respuesta :

MathPhys

Answer:

400 J

Explanation:

Given:

Δy = 4.00 m

v₀ = 0 m/s

a = 5.00 m/s²

Find: v²

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (5.00 m/s²) (4.00 m)

v² = 40.0 m²/s²

Find KE:

KE = ½ mv²

KE = ½ (20.0 kg) (40.0 m²/s²)

KE = 400 J

What is Kinetic energy ?

The kinetic energy of the object is the energy is that energy which possesses due to the its motion. It is the work needed to accelerate a body of a given mass from rest to its stated velocity .

Kinetic energy (KE) = 1/2 *m* v^2

mass = 20 kg (given )

accelerates = a = 5  m/s^2

to find = v ?

u = 0

using newton's kinematics equation  

v^2 - u^2 = 2as

v^2 - 0 = 2 * (5) * (4)

v^2 =  40

v = [tex]\sqrt{40}[/tex] m/s

Kinetic energy (KE) = 1/2 *m* v^2

                                =  1/2 * (20 kg) * 40 m/s

                                = 400 J

The kinetic energy (in J) of the crate when it is 4.0 m above the floor will be 400 J

learn more about  kinetic energy

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