Question

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. Be sure to show your work for both approaches. [Use 9.81 m/s2 for g and assume the barbell was at rest right before it was dropped.] Vertical velocity of barbell at 0.21 m ______________ Vertical velocity right before the barbell hits the ground ______________

Answer 1

Total energy = Maximum potential energy = 1256.66 J

At a height of h = 0.21 m, potential energy is:

U = mgh = 305 x 9.81 x 0.21 = 628.33 J

so, kinetic energy at height h = 0.21 m is:

K = 1256.66 - 628.33 = 628.33 J

But, K = (1/2)mv²

=> v = 2.03 m/s

this is the magnitude of vertical velocity of the barbell when it reaches 0.21 m.

When the barbell reaches the ground, Total energy = maximum kinetic energy

=> (1/2)mv² = 1256.66

=> v = 2.87 m/s

Other approach:

u = 0 m/s, a = 9.81m/s² and h = 0.21 m

use v² = u² + 2ah

=> v = 2.03 m/s.

for h = 0.42m,

v² = 0 + 2(9.81)(0.42)

=> v = 2.87 m/s.