A male cheerleader does 2.25 kJ of work in lifting a 57.5 kg cheerleader 1.25 m off the ground during the half time show. The energy gained by the lifted cheerleader is __________ kJ.
force * distance in direction of force
= 57.5 * 9.81 * 1.25
= 705 Joules = 0.705 kJ
the rest of the 2.25 kJ is lost to heat and his own motion
To find the energy gained by the lifted cheerleader, we need to use the work-energy principle. The work done on an object is equal to the change in its potential energy.
The formula for work is given by W = F * d * cosθ, where W is the work done, F is the force applied, d is the displacement, and θ is the angle between the force and displacement vectors.
In this case, the work done is given as 2.25 kJ (kilojoules) and the displacement is 1.25 m.
Since we are lifting the cheerleader against gravity, the force applied is equal to the weight of the cheerleader, which can be calculated as F = m * g, where m is the mass of the cheerleader and g is the acceleration due to gravity (approximately 9.8 m/s²).
Let's calculate the force applied first:
F = m * g
F = 57.5 kg * 9.8 m/s²
F ≈ 563.5 N
Now we can calculate the energy gained by the lifted cheerleader:
W = F * d * cosθ
W = 563.5 N * 1.25 m * cos(0°) [since the force and displacement are in the same direction, cos(0°) = 1]
W = 563.5 N * 1.25 m
W = 704.38 J (joules)
To convert the energy from joules to kilojoules, we divide by 1000:
W = 704.38 J ÷ 1000
W ≈ 0.704 kJ
Therefore, the energy gained by the lifted cheerleader is approximately 0.704 kJ.