A power lifter with a body mass of 130 kg performs a “dead lift”, raising a barb

Question

A power lifter with a body mass of 130 kg performs a “dead lift”, raising a barbell with a mass of 350 kg to a height of 0.46 m above the ground. How much mechanical work has he done on the barbell? At the top of the lift, what is the potential energy of the barbell relative to the ground? If the power lifter drops the barbell, what will the magnitude of its vertical velocity be when it is 0.23 m above the ground? What will the magnitude of its vertical velocity be immediately before hitting the ground? [Use a mechanical energy approach to answer the last two questions, not a projectile motion approach.]

Work performed on barbell to get it to top of lift (0.46 m)______________

Potential energy of barbell at top of lift (0.46 m)______________

Velocity at 0.23 m______________

Velocity immediately before hitting the ground______________

Explanation / Answer

a) Work done on the barbell to get it to top of lift = Chnage in PE at the top = mgh = 350 * 9.8 * 0.46 = 1577.8 N

b) Potential energy of barbell at top of lift = mgh = 1577.8N

c) From top(=0.46m) to at 0.23 m above the ground,

Loss in PE = Gain in KE

mg(0.46 - 0.23) = 0.5*m*(v2-0)

=> v = 2.12m/s

d) Loss in PE = Gain in KE

mgh = 0.5*m*(v2-0)

=> v = 3m/s

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