A power lifter performs a dead lift, raising a barbell with a mass of 305 kg to a height of 0.42 m above the ground, giving the barbell 1256.66 J of potential energy. The power lifter then releases the barbell, letting it drop towards the ground. Determine the magnitude of the vertical velocity of the barbell when it reaches a height of 0.21 m (on the way down) using a mechanical energy approach. Then calculate the velocity right as it reaches the ground using a mechanical energy approach. Confirm your answer for the velocity of the barbell right before it hits the ground by also calculating this velocity using a projectile motion approach.

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MathPhys MathPhys · Answer 1 · 2019-05-13T15:45:43+02:00

Answer:

Explanation:

Before it hits the ground:

The initial potential energy = the final potential energy + the kinetic energy

mgH = mgh + 1/2 mv²

gH = gh + 1/2 v²

v = √(2g (H - h))

v = √(2 * 9.81 m/s² * (0.42 m - 0.21 m))

v ≈ 2.0 m/s

When it hits the ground:

Initial potential energy = final kinetic energy

mgH = 1/2 mv²

v = √(2gH)

v = √(2 * 9.81 m/s² * 0.42 m)

v ≈ 2.9 m/s

Using a kinematic equation to check our answer:

v² = v₀² + 2a(x - x₀)

v² = (0 m/s)² + 2(9.8 m/s²)(0.42 m)

v ≈ 2.9 m/s