Work Done by Theme Park Machine on Rider

In summary: 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
Balsam
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.
 

Related to Work Done by Theme Park Machine on Rider

1. What is work done by a theme park machine on a rider?

Work done by a theme park machine on a rider refers to the amount of energy transferred from the machine to the rider while they are on a ride. This energy is typically in the form of mechanical work, which results in the rider experiencing forces and motion during the ride.

2. How is work done by a theme park machine on a rider calculated?

The work done by a theme park machine on a rider can be calculated using the formula W = Fd, where W is work, F is the force applied by the machine, and d is the distance the rider travels on the ride. This calculation is based on the principle of work, which states that work is equal to the force applied multiplied by the distance moved in the direction of the force.

3. What factors affect the amount of work done by a theme park machine on a rider?

The amount of work done by a theme park machine on a rider can be affected by several factors, including the type of ride, the speed and acceleration of the ride, the weight and size of the rider, and the duration of the ride. Additionally, external factors such as friction and air resistance can also impact the amount of work done.

4. Is work done by a theme park machine on a rider a measure of safety?

No, work done by a theme park machine on a rider is not a measure of safety. While the amount of work done can indicate the level of force and motion experienced by the rider, it does not take into account other safety factors such as proper restraint systems, maintenance of the ride, and adherence to safety regulations.

5. Can the work done by a theme park machine on a rider be negative?

Yes, the work done by a theme park machine on a rider can be negative if the direction of the force applied by the machine is opposite to the direction of the rider's motion. This can occur on certain rides where the machine is used to slow down or stop the rider's motion, resulting in negative work being done on the rider.

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