PE=mgh=12*9.8*2.5 joules
work done: same
power= workdone/seconds
An initially stationary 15.0-kg crate of cheese wheels is pulled via a cable a distance d= 5.70m up a frictionless ramp to a height of h= 2.50m, where it stops. What is the potential energy of the crate after it is lifted? How much work is done during the lift? If it takes 10 seconds to lift the cheese, what is the power rating of the lifting device?
3 answers
PE= 294
work done= 294
power= 29.4
work done= 294
power= 29.4
potential energy = mass * gravity * height
= 15.0kg * 9.8 m/s^2 * 2.50 m
= 367.875 J or 368 J (sig figs)
work = mass * gravity (OR " force ") * distance
= 15.0 kg * 9.8 m/s^2 * 5.70 m
= 837.9 J or 838 J (sig figs)
power = work / time
= 837.9 J / 10 s
= 83.79 J/s or 83.8 J/s (sig figs)
= 15.0kg * 9.8 m/s^2 * 2.50 m
= 367.875 J or 368 J (sig figs)
work = mass * gravity (OR " force ") * distance
= 15.0 kg * 9.8 m/s^2 * 5.70 m
= 837.9 J or 838 J (sig figs)
power = work / time
= 837.9 J / 10 s
= 83.79 J/s or 83.8 J/s (sig figs)