Aweightlifter presses a 400 n weight 0.5 m over his head in 2 seconds, what is the power of the weight lifter

There are two ways to solve this problem. First we write the given.

Given: Force F = 400 N;  Height h = 0.5 m;  Time t = 2 s

Formula: P = W/t;  but Work W = Force x distance or W = f x d
 
Weight is also a Force, therefore:  W = mg, solve for Mass m = ?

m = w/g  m = 400 N/9.8 m/s²  m = 40.82 Kg

P = W/t = F x d/t  = mgh/t  P = (40.82 Kg)(9.8 m/s²)/2 s  

P = 100 J/s or 100 Watts

Explanation:

This problem is based on conservation of rotational momentum.

Moment of inertia of rod about its center

= 1/12 m l² , m is mass of the rod and l is its length .

= 1 / 12 x 4.6 x .11²

I = .004638 kg m²

The angular momentum of the bullet about the center of rod = mvr

where m is mass , v is perpendicular component of velocity of bullet and r is distance of point of impact of bullet fro center .

5 x 10⁻³ x v sin60 x .11 x .5  where v is velocity of bullet

According to law of conservation of angular momentum

5 x 10⁻³ x v sin60 x .11 x .5 = ( I + mr²)ω , where ω is angular velocity of bullet rod system and  ( I + mr²) is moment of inertia of bullet rod system .

.238 x 10⁻³ v = ( .004638 + 5 x 10⁻³ x .11² x .5² ) x 12

.238 x 10⁻³ v = ( .004638 + .000015125 ) x 12

.238 x 10⁻³ v = 55.8375 x 10⁻³

.238 v = 55.8375

v = 234.6 m /s