/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 A \(70-\mathrm{kg}\) person ride... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(70-\mathrm{kg}\) person rides in a \(30-\mathrm{kg}\) cart moving at 12 \(\mathrm{m} / \mathrm{s}\) at the top of a hill that is in the shape of an are of a circle with a radius of 40 \(\mathrm{m}\) . (a) What is the apparent weight of the person as the cart passes over the top of the hill? (b) Determine the maximum speed that the cart may travel at the top of the hill without losing contact with the surface. Does your answer depend on the mass of the cart or the mass of the person? Explain.

Short Answer

Expert verified
(a) Apparent weight is 434 N. (b) Maximum speed is 19.8 m/s; it does not depend on the mass.

Step by step solution

01

Understanding the scenario

We have a 70 kg person in a 30 kg cart, moving with a velocity of 12 m/s at the top of a circular hill. The hill has a radius of 40 m. First, we identify the forces acting on the person at the top of the hill: gravitational force directed downwards and normal force (apparent weight) acting upwards.
02

Determine gravitational force

Calculate the gravitational force acting on the person. The gravitational force is given by:\[ F_{gravity} = m_{person} \cdot g = 70 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 686 \, \text{N}. \]
03

Calculate centripetal force at the top of the hill

Since the motion is circular, a centripetal force is required. The centripetal force is provided by the difference between the gravitational force and the normal force:\[ F_{centripetal} = \frac{mv^2}{r}, \]where \( m \) is the mass of the person, \( v \) is the velocity, and \( r \) is the radius of the hill.
04

Calculate the centripetal force experienced by the person

Plug in the values:\[ F_{centripetal} = \frac{70 \, \text{kg} \times (12 \, \text{m/s})^2}{40 \, \text{m}} = 252 \, \text{N}. \]
05

Determine the apparent weight

The apparent weight is the normal force, \( F_{normal} \), which is the gravitational force minus the centripetal force:\[ F_{normal} = F_{gravity} - F_{centripetal} = 686 \, \text{N} - 252 \, \text{N} = 434 \, \text{N}. \]
06

Calculate maximum speed without losing contact

To prevent the cart from losing contact, the apparent weight (normal force) should remain at least zero. Set apparent weight as zero and solve for \( v \):\[ F_{gravity} = F_{centripetal} \Rightarrow 686 \, \text{N} = \frac{70 \, \text{kg} \times v^2}{40 \, \text{m}}. \]
07

Solve for critical speed

Rearranging and solving for \( v \):\[ v^2 = \frac{686 \, \text{N} \times 40 \, \text{m}}{70 \, \text{kg}} \Rightarrow v = \sqrt{392} = 19.8 \, \text{m/s}. \]
08

Understand mass dependence

Notice that the mass of the cart is not part of the equation involving speed and critical points. In circular motion under gravity, the critical speed depends solely on the gravitational force and the radius of the circle, not on the mass of the cart or person.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Centripetal Force
Centripetal force is the inward force required to keep an object moving in a circular path. In our scenario, this force is critical at the top of the hill to maintain circular motion. When an object travels in a circle, it naturally wants to move outward due to inertia. However, to keep it from flying off the path, a force directed towards the center is necessary.
  • Mathematically, centripetal force is expressed as: \[ F_{centripetal} = \frac{mv^2}{r} \ \text{where } m \text{ is mass, } v \text{ is velocity, and } r \text{ is the radius of the circle.} \]
  • This force ensures that the object remains on its curved trajectory.
In the case of our 70 kg person, the centripetal force at the top of the hill is the difference between gravitational force and the apparent weight. This balance of forces ensures that the person stays safely in the cart without flying off the hill. By understanding and calculating this force, we can determine how fast the cart can move without losing contact with the hill's surface.
Circular Motion
Circular motion is a type of movement wherein an object moves in a circular path. This motion can be uniform (constant speed) or non-uniform (changing speed), but always requires an inward (centripetal) force to sustain the motion.
  • In our example, the cart and person are at the top of a circular hill, and their combined motion is indicative of uniform circular motion.
  • The velocity at which they move influences the apparent weight, or how heavy the person seems due to the interaction between gravity and centrifugal tendencies.
Importantly, when discussing circular motion, the concept of apparent weight is observed. At the top of the hill, the forces of gravity, inertia, and normal (apparent weight) interact. As velocity increases towards critical speed, the apparent weight approaches zero, alerting us to the edge of losing contact with the path.
Gravitational Force
Gravitational force is the attractive force between two masses, in this context, between the Earth and the person in the cart. This force acts downwards, and is crucial in determining how objects move on Earth.
  • The equation for gravitational force is: \[ F_{gravity} = m \cdot g \ \text{where } m \text{ is mass and } g \text{ is the acceleration due to gravity (} 9.8 \, \text{m/s}^2\text{).} \]
  • This force is a constant downwards pull that provides the necessary energy for centripetal force in circular motion.
In our problem, gravitational force acts directly inward towards the hill's center, which counteracts the outward forces from circular motion. It's essential for calculating apparent weight, especially when determining whether the cart can maintain contact with the entire circular path of the hill. By comparing gravitational force to the centripetal force, we can find the critical speed at which the apparent weight becomes crucial for staying on course.

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