/*! 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 25 A \(5.0\)-kg bucket of water is ... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(5.0\)-kg bucket of water is raised from a well by a rope. If the upward acceleration of the bucket is \(3.0 \mathrm{~m} / \mathrm{s}^{2}\), find the force exerted by the rope on the bucket.

Short Answer

Expert verified
The force exerted by the rope on the bucket is \(64 \, N\).

Step by step solution

01

Identify Given Parameters and Formula

The problem provides a mass \( m = 5.0 \) kg and an upward acceleration \( a = 3.0 \, m/s^{2} \). We will use these values in the equation \( F=ma \) to find the force exerted by the rope. Additionally, we need to consider the force due to gravity, which is \( F_{g}=m \cdot g \), where \( g = 9.8 \, m/s^{2} \) (acceleration due to gravity).
02

Calculate the Force due to Gravity

First let's calculate the force acting on the bucket due to gravity: \( F_{g}= m \cdot g = 5.0 \, kg \cdot 9.8 \, m/s^{2} = 49 \, N \). This is the force pulling the bucket downwards.
03

Calculate the Upward Force exerted by the Rope

The force exerted by the rope pulling the bucket upwards (without considering gravity) is given by \( F_{ext}=m \cdot a = 5.0 \, kg \cdot 3.0 \, m/s^{2} = 15 \, N \).
04

Calculate the Net Upward Force exerted by the Rope

The actual force exerted by the rope must overcome the force of gravity. Therefore, the net force, \( F_{net} \), is given by \( F_{net} = F_{ext} + F_{g} = 15 \, N + 49 \, N = 64 \, N \).
05

Conclusion

Therefore, the force exerted by the rope on the bucket to lift it with an upward acceleration of \( 3.0 \, m/s^{2} \) is 64 N.

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Key Concepts

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

Force Calculation
The essence of force calculation lies in understanding Newton's Second Law of Motion. This law can be neatly encapsulated in the equation \( F = ma \), where \( F \) represents the force applied, \( m \) is the mass of the object, and \( a \) is the acceleration. The formula suggests that an object will only accelerate if a net force is acting upon it. The greater the mass, the more force required to achieve the same acceleration. Conversely, with a fixed mass, greater acceleration demands a higher force.
This concept is beautifully illustrated with the bucket example. We have a bucket that weighs \(5.0\,\text{kg}\) and undergoes an acceleration of \(3.0\,\text{m/s}^2\). Through the use of Newton's formula, the force applied by the rope to achieve this acceleration can initially be calculated as \( F_{ext} = m \cdot a = 5.0\,\text{kg} \cdot 3.0\,\text{m/s}^2 = 15\,\text{N} \).
This initial calculation considers only the acceleration without accounting for additional forces such as gravity, which have significant effects in real-world situations.
Acceleration
Acceleration serves as a fundamental concept in physics, denoting the rate of change of velocity over time. Simply put, it's how quickly an object speeds up or slows down. When an object is subject to a force, particularly a net force, acceleration is the outcome.
In our example, the bucket experiences an acceleration due to the upward force applied via the rope. The rope generates an upward acceleration of \(3.0\,\text{m/s}^2\), a clear indicator of the force applied overcoming both the bucket's mass and any opposing forces like gravity.
  • Acceleration is measured in meters per second squared (\(\text{m/s}^2\)).
  • It can be either positive or negative, depending on the direction relative to the initial motion.
It's imperative to note that acceleration is not solely about speed increments. It embodies any velocity change, inclusive of deceleration or direction shifts. Thus, understanding acceleration broadens as we consider its role in altering motion through the applied forces in diverse contexts.
Gravity
Gravity is a ubiquitous and constant force acting on objects by virtue of their mass and Earth's gravitational pull. Its role is pivotal in scenarios involving vertical motion, such as when lifting objects against gravity, thereby creating scenarios where calculations must adjust for its influence.
In the context of the exercise with the bucket, gravity applies a downward force calculated through the formula \( F_g = m \cdot g \), where \( m \) is the mass and \( g \) is the acceleration due to gravity, approximately \( 9.8\,\text{m/s}^2 \). For the bucket weighing \(5.0\,\text{kg}\), the gravitational force amounts to \(49\,\text{N}\). This force acts downward, opposing the upward force exerted by the rope.
  • Earth's gravity typically accelerates objects at approximately \(9.8\,\text{m/s}^2\).
  • Gravity's force is universal, influencing all masses regardless of their size.
In any force calculation involving Earth's surface, accounting for gravity's effect is necessary unless conditions imply otherwise, such as in a vacuum or space.

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Most popular questions from this chapter

A \(5.0-g\) bullet leaves the muzzle of a rifle with a speed of \(320 \mathrm{~m} / \mathrm{s}\). What force (assumed constant) is exerted on the bullet while it is traveling down the \(0.82\)-m-long barrel of the rifle?

A fisherman poles a boat as he searches for his next catch. He pushes parallel to the length of the light pole, exerting a force of \(240 \mathrm{~N}\) on the bottom of a shallow lake. The pole lies in the vertical plane containing the boat's keel. At one moment, the pole makes an angle of \(35.0^{\circ}\) with the vertical and the water exerts a horizontal drag force of \(47.5 \mathrm{~N}\) on the boat, opposite to its forward velocity of magnitude \(0.857 \mathrm{~m} / \mathrm{s}\). The mass of the boat including its cargo and the worker is \(370 \mathrm{~kg}\). (a) The water exerts a buoyant force vertically upward on the boat. Find the magnitude of this force. (b) Assume the forces are constant over a short interval of time. Find the velocity of the boat \(0.450 \mathrm{~s}\) after the moment described. (c) If the angle of the pole with respect to the vertical increased but the exerted force against the bottom remained the same, what would happen to buoyant force and the acceleration of the boat?

After falling from rest from a height of \(30 \mathrm{~m}\), a \(0.50-\mathrm{kg}\) ball rebounds upward, reaching a height of \(20 \mathrm{~m}\). If the contact between ball and ground lasted \(2.0 \mathrm{~ms}\), what average force was exerted on the ball?

A student decides to move a box of books into her dormitory room by pulling on a rope attached to the box. She pulls with a force of \(80.0 \mathrm{~N}\) at an angle of \(25.0^{\circ}\) above the horizontal. The box has a mass of \(25.0\) \(\mathrm{kg}\), and the coefficient of kinetic friction between box and floor is \(0.300\). (a) Find the acceleration of the box. (b) The student now starts moving the box up a \(10.0^{\circ}\) incline, keeping her \(80.0 \mathrm{~N}\) force directed at \(25.0^{\circ}\) above the line of the incline. If the coefficient of friction is unchanged, what is the new acceleration of the box?

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force \(F\) exerted by the tail fin and the downward force due to gravity. A record Chinook salmon has a length of \(1.50 \mathrm{~m}\) and a mass of \(61.0 \mathrm{~kg}\). If this fish is moving upward at \(3.00 \mathrm{~m} / \mathrm{s}\) as its head first breaks the surface and has an upward speed of \(6.00 \mathrm{~m} / \mathrm{s}\) after two-thirds of its length has left the surface, assume constant acceleration and determine (a) the salmon's acceleration and (b) the magnitude of the force \(F\) during this interval.

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