A cheerleader lifts his 77.6 kg partner straight up off the ground a distance of 0.574 m before releasing her. Assume the partner's velocity is zero at the beginning and the end of the lift.

The acceleration of gravity is 9.8 m/s?
If he does this 29 times, how much work has he done?
Answer in units of J. Answer in units of J.

1 answer

To calculate the work done by the cheerleader in lifting his partner, we can use the formula for gravitational potential energy, which is related to work. The work done \( W \) in lifting an object is given by:

\[ W = m \cdot g \cdot h \]

where:

  • \( W \) is the work done,
  • \( m \) is the mass of the partner (in kg),
  • \( g \) is the acceleration due to gravity (in m/s²),
  • \( h \) is the height lifted (in m).

Given:

  • \( m = 77.6 \) kg,
  • \( g = 9.8 \) m/s²,
  • \( h = 0.574 \) m.

First, we'll calculate the work done for one lift:

\[ W_{single} = 77.6 , \text{kg} \cdot 9.8 , \text{m/s}^2 \cdot 0.574 , \text{m} \]

Now, let's calculate that step-by-step:

  1. Calculate \( m \cdot g \):

\[ m \cdot g = 77.6 , \text{kg} \cdot 9.8 , \text{m/s}^2 = 760.48 , \text{N} \]

  1. Now multiply by the height \( h \):

\[ W_{single} = 760.48 , \text{N} \cdot 0.574 , \text{m} \approx 436.29 , \text{J} \]

Now, since he does this 29 times, the total work done \( W_{total} \) will be:

\[ W_{total} = W_{single} \cdot \text{number of lifts} = 436.29 , \text{J} \cdot 29 \]

Calculating this:

\[ W_{total} \approx 436.29 , \text{J} \cdot 29 \approx 12656.41 , \text{J} \]

Therefore, the total work done by the cheerleader in lifting his partner 29 times is approximately:

\[ \boxed{12656.41 , \text{J}} \]