Problem 31
Question
A 5.60-kg bucket of water is accelerated upward by a cord of negligible mass whose breaking strength is 75.0 N. If the bucket starts from rest, what is the minimum time required to raise the bucket a vertical distance of 12.0 m without breaking the cord?
Step-by-Step Solution
Verified Answer
The minimum time required is approximately 2.58 seconds.
1Step 1: Identify the forces involved
Determine the forces acting on the bucket: the gravitational force (weight) and the tension force in the cord. The gravitational force is given by the formula: \( F_g = m \cdot g \), where \( m = 5.60 \text{ kg} \) is the mass of the bucket, and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity.
2Step 2: Calculate the gravitational force
Using the formula \( F_g = m \cdot g \), we find the gravitational force: \( F_g = 5.60 \times 9.8 = 54.88 \text{ N} \).
3Step 3: Determine maximum tension for acceleration
Since the breaking strength of the cord is 75.0 N, the maximum net force for safe acceleration is \( 75.0 \text{ N} - 54.88 \text{ N} = 20.12 \text{ N} \). This is the maximum upward force before the cord breaks.
4Step 4: Apply Newton's Second Law
Use \( F = m \cdot a \) to find the maximum acceleration the bucket can have without breaking the cord: \( 20.12 = 5.60 \cdot a \). Solving for \( a \), we get \( a = \frac{20.12}{5.60} = 3.59 \text{ m/s}^2 \).
5Step 5: Calculate the minimum time for the vertical motion
Use the kinematic equation for motion: \( d = \frac{1}{2} a t^2 \), where \( d = 12.0 \text{ m} \). Substituting the values, we get \( 12.0 = \frac{1}{2} \times 3.59 \times t^2 \). Simplify to find \( t^2 = \frac{12.0}{1.795} \) which results in \( t^2 = 6.68 \). Find \( t \): \( t = \sqrt{6.68} \approx 2.58 \text{ seconds} \).
Key Concepts
Gravitational ForceKinematics EquationsMechanics Problems
Gravitational Force
Gravitational force is a fundamental force of nature that acts between two masses. This is the force that makes apples fall from trees and keeps planets in orbit around the sun. In mechanics problems, especially those involving objects in motion, understanding gravitational force is crucial. Gravitational force on an object near the Earth's surface is calculated using the equation: \[ F_g = m \cdot g \] where:
- \( F_g \) is the gravitational force or weight.
- \( m \) is the mass of the object, measured in kilograms (kg).
- \( g \) is the acceleration due to gravity, approximately \( 9.8 \, \text{m/s}^2 \) on Earth.
Kinematics Equations
Kinematics equations are used to describe the motion of objects. They allow us to calculate unknown variables, such as time, distance, acceleration, and velocity, when certain conditions are known. When dealing with vertical motion and constant acceleration in mechanics problems, the relevant kinematic equation is: \[ d = \frac{1}{2} a t^2 \] where:
- \( d \) is the distance traveled, measured in meters (m).
- \( a \) is acceleration, in meters per second squared (\( \text{m/s}^2 \)).
- \( t \) is time, in seconds (s).
Mechanics Problems
Mechanics problems involve understanding how forces affect the motion of objects, using principles laid down by Sir Isaac Newton. Solving these problems typically requires:
- Identifying all forces acting on an object, in this case, gravitational force and tension.
- Applying Newton's Second Law, \( F = m \cdot a \), to relate forces to acceleration.
- Using kinematics to connect acceleration, distance, and time.
Other exercises in this chapter
Problem 29
A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force \(F =\) 40.0 N that is directed at
View solution Problem 30
A large box containing your new computer sits on the bed of your pickup truck. You are stopped at a red light. When the light turns green, you stomp on the gas
View solution Problem 32
You have just landed on Planet X. You release a 100-g ball from rest from a height of 10.0 m and measure that it takes 3.40 s to reach the ground. Ignore any fo
View solution Problem 35
Basketball player Darrell Griffith is on record as attaining a standing vertical jump of 1.2 m (4 ft). (This means that he moved upward by 1.2 m after his feet
View solution