Work Done by Theme Park Machine on Rider

The normal force (Fn) is equal to the gravitational force (Fg=mg) in this case, as the rider is not accelerating while being lifted at a constant speed. Therefore, the work done on the rider by the machine is equal to the force exerted (Fg=mg) multiplied by the displacement (78m) and the cosine of the angle between the force and the displacement (θ=0). This results in a work done of 4,368 Joules.
  • #1
226
8

Homework Statement


A drop tower lifts riders at a constant speed to a height of 78m and suddenly drops them. Determine the work done on a 56kg rider by the machine as she is lifted to the top of the ride.
F=mg
displacement=78m
θ= 0 degrees

Homework Equations


W=FΔdcosθ

The Attempt at a Solution


I solved this problem, by plugging in mg as the force. But, I am confused. If you plug in mg as the force exerted by the machine on the rider, doesn't that mean that the normal force exerted on the rider is the force that lifts them up 78m. How does that make any sense? Shouldn't there be a force with a magnitude greater than Fg to lift the rider up. Doesn't having Fn and Fg as the only y component forces mean that there's no (Fnet)y? That would mean there's no acceleration in the y component, which there is because the rider is lifted up 78m. Can someone please explain this to me?
 
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  • #2
Fnet = T - mg = ma
Where; T = tension (at rest).

The problem statement seems to be only concerned with the amount of acceleration force needed to lift the '56kg rider'.

Can we presume that the machine is designed to hold multiple riders even when turned off, such that energy is only consumed through acceleration?
 
  • #3
Balsam said:

Homework Statement


A drop tower lifts riders at a constant speed to a height of 78m and suddenly drops them. Determine the work done on a 56kg rider by the machine as she is lifted to the top of the ride.
F=mg
displacement=78m
θ= 0 degrees

Homework Equations


W=FΔdcosθ

The Attempt at a Solution


I solved this problem, by plugging in mg as the force. But, I am confused. If you plug in mg as the force exerted by the machine on the rider, doesn't that mean that the normal force exerted on the rider is the force that lifts them up 78m. How does that make any sense? Shouldn't there be a force with a magnitude greater than Fg to lift the rider up. Doesn't having Fn and Fg as the only y component forces mean that there's no (Fnet)y? That would mean there's no acceleration in the y component, which there is because the rider is lifted up 78m. Can someone please explain this to me?
A force only slightly higher than mg is all that is required to lift the rider.
 

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