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Chapter 6: Problem 33

How large a force is required to accelerate a 1300 -kg car from rest to a speed of \(20 \mathrm{~m} / \mathrm{s}\) in a horizontal distance of \(80 \mathrm{~m}\)?

Short Answer

Expert verified
The required force is 3250 N.

Step by step solution

01

Identify Given Values

Identify the known values from the problem. We have the mass of the car \( m = 1300 \, \text{kg} \), initial velocity \( u = 0 \, \text{m/s} \) (since the car starts from rest), final velocity \( v = 20 \, \text{m/s} \), and the distance \( s = 80 \, \text{m} \).
02

Use Kinematic Equation

Apply the kinematic equation to find the acceleration \( a \). The equation is \( v^2 = u^2 + 2as \). Substitute the values: \( 20^2 = 0 + 2a(80) \).
03

Solve for Acceleration

Solve the equation from the previous step: \( 400 = 160a \). So, \( a = \frac{400}{160} = 2.5 \, \text{m/s}^2 \).
04

Apply Newton's Second Law

Use Newton's second law \( F = ma \) to find the force \( F \). Substitute the known values: \( F = 1300 \, \text{kg} \times 2.5 \, \text{m/s}^2 \).
05

Calculate the Force

Perform the calculation from the previous step: \( F = 1300 \times 2.5 = 3250 \, \text{N} \).

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Key Concepts

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

Kinematics
Kinematics is the branch of physics that studies the motion of objects without considering the causes of this motion. It helps us understand how objects move and use mathematical equations to describe different aspects of motion. In this context, we are interested in the relationship between displacement, velocity, and acceleration.
The primary kinematic equation used here is:
  • \( v^2 = u^2 + 2as \)
Here, \( v \) is the final velocity of the object, \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the displacement. With this formula, we can solve for one unknown if we have the other values.
By rearranging this equation, we can find out how the car in the example moves from rest to a speed of 20 m/s over a distance of 80 m. Knowing how to use this equation is crucial in solving kinematic problems, making it a vital part of understanding physics in general.
Acceleration Calculations
Acceleration describes how quickly an object's velocity changes over time. It's given in meters per second squared (m/s虏) and can be calculated by rearranging kinematic equations.
In the exercise, to determine the car's acceleration, we used the kinematic equation:
  • First, write the equation: \( v^2 = u^2 + 2as \)
  • Then, plug in the values: \( 20^2 = 0 + 2a(80) \)
  • Next, solve for \( a \) by simplifying the equation to \( 400 = 160a \)
  • Finally, solve for acceleration: \( a = \frac{400}{160} = 2.5 \, \text{m/s}^2 \)
Finding the acceleration is key for determining the force required, as it tells us how much the car's speed increases per unit of time.
Force Calculation
Calculating force involves understanding Newton's Second Law, which states: \( F = ma \). This equation tells us that the force exerted on an object equals the mass of the object multiplied by its acceleration.
In the given example, after determining the car's acceleration:
  • You know the mass \( m = 1300 \, \text{kg} \)
  • You know the acceleration \( a = 2.5 \, \text{m/s}^2 \)
  • From Newton's Second Law, calculate the force: \( F = 1300 \, \text{kg} \times 2.5 \, \text{m/s}^2 \)
  • Finally, the force needed is \( F = 3250 \, \text{N} \)
Understanding this calculation is essential, as it connects the concepts of mass and acceleration to produce a measurable force, allowing us to predict the behavior of objects under different conditions.

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Most popular questions from this chapter

A 0.50-kg block slides across a tabletop with an initial velocity of \(20 \mathrm{~cm} / \mathrm{s}\) and comes to rest in a distance of \(70 \mathrm{~cm}\). Find the average friction force that retarded its motion. The KE of the block is decreased because of the slowing action of the friction force. That is, Change in KE of block = Work done on block by friction force $$ \frac{1}{2} m v_{f}^{2}-\frac{1}{2} m v_{i}^{2}=F_{f} s \cos \theta $$ Because the friction force on the block is opposite in direction to the displacement, \(\cos \theta=-1\). Using \(v_{f}=0, v_{i}=0.20 \mathrm{~m} / \mathrm{s}\), and \(s=\) \(0.70 \mathrm{~m}\), the above equation becomes $$ 0-\frac{1}{2}(0.50 \mathrm{~kg})(0.20 \mathrm{~m} / \mathrm{s})^{2}=\left(F_{\mathrm{f}}\right)(0.70 \mathrm{~m})(-1) $$ from which \(F_{\mathrm{f}}=0.014 \mathrm{~N}\).

In Fig. 6-1, assume that the object is being pulled in a straight line along the ground by a 75-N force directed \(28^{\circ}\) above the horizontal. How much work does the force do in pulling the object \(8.0 \mathrm{~m}\) horizontally? The work done is equal to the product of the displacement, \(8.0 \mathrm{~m}\), and the component of the force that is parallel to the displacement, \((75 \mathrm{~N})\left(\cos 28^{\circ}\right)\). Thus, $$ W=(75 \mathrm{~N})\left(\cos 28^{\circ}\right)(8.0 \mathrm{~m})=0.53 \mathrm{~kJ} $$

A \(0.50\) -kg ball falls past a window that is \(1.50 \mathrm{~m}\) in vertical length. (a) How much did the KE of the ball increase as it fell past the window? \((b)\) If its speed was \(3.0 \mathrm{~m} / \mathrm{s}\) at the top of the window, what was its speed at the bottom?

A moving 300 -g object slides unpushed \(80 \mathrm{~cm}\) in a straight line along a horizontal tabletop. How much work is done in overcoming friction between the object and the table if the coefficient of kinetic friction is \(0.20\) ? First find the friction force. Since the normal force equals the weight of the object, $$ F_{\mathrm{f}}=\mu_{k} F_{N}=(0.20)(0.300 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=0.588 \mathrm{~N} $$ The work done overcoming friction is \(F_{\mathrm{f}} S \cos \theta\). Here \(\theta\) is the angle between the force and the displacement. Because the friction force is opposite in direction to the displacement, \(\theta=180^{\circ}\). Therefore, Work \(=F_{\mathrm{f}} S \cos 180^{\circ}=(0.588 \mathrm{~N})(0.80 \mathrm{~m})(-1)=-0.47 \mathrm{~J}\) The work is negative because the friction force is oppositely directed to the displacement; it slows the object and it decreases the object's kinetic energy, or more to the point, it opposes the motion.

Imagine a 60.0-kg skier standing still on the top of a snow-covered hill \(150 \mathrm{~m}\) high. Neglecting any friction losses, how fast will she be moving at the bottom of the hill? Does her mass matter?
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