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A \(423.5-\mathrm{N}\) force accelerates a go-cart and its driver from \(10.4 \mathrm{~m} / \mathrm{s}\) to \(17.9 \mathrm{~m} / \mathrm{s}\) in \(5.00 \mathrm{~s}\). What is the mass of the go-cart plus driver?

Short Answer

Expert verified
Answer: The mass of the go-cart plus the driver is 282.33 kg.

Step by step solution

01

Calculate the acceleration

To find the acceleration, we'll use the formula: Acceleration = (Final velocity - Initial velocity) / Time Given: Initial velocity (u) = 10.4 m/s Final velocity (v) = 17.9 m/s Time (t) = 5.00 s Now, let's calculate the acceleration (a): a = (17.9 - 10.4) / 5.00 a = 7.5 / 5.00 a = 1.50 m/s²
02

Use Newton's second law to find the mass

Newton's second law states that: Force (F) = Mass (m) × Acceleration (a) The problem gives us the force F = 423.5 N and we have calculated the acceleration, a = 1.50 m/s². Now we can solve for the mass (m): m = F / a m = 423.5 N / 1.50 m/s² m = 282.33 kg
03

Write the final answer

The mass of the go-cart plus the driver is 282.33 kg.

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Key Concepts

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

Understanding Force Calculation
Let's start by exploring the concept of force calculation within the realm of physics. Force is perhaps one of the most fundamental aspects when it comes to understanding motion. The unit of force in the International System of Units (SI) is the Newton (N), named after Sir Isaac Newton, who proposed the laws of motion.

Force can be thought of as a push or pull acting on an object and is calculated using Newton's second law of motion, which states that force equals mass times acceleration (\( F = m \times a \) ). When you're given the acceleration and the mass of an object, like in our go-cart scenario, simply multiply them to find the force exerted. Conversely, if you know the force and acceleration, you can rearrange the formula to find the mass, which is the process that was used in the textbook problem.

It's important to note that forces can act in various directions and can be the sum of multiple individual forces acting on the same object. In our exercise scenario, the force given is the net force acting to accelerate the go-cart and its driver. This concept is crucial for correctly applying the formula in different contexts.
The Acceleration Formula Explained
Acceleration is the rate at which an object's velocity changes over time. In the realms of physics, we often encounter problems where we need to calculate this vital quantity. The basic acceleration formula, which can be easily remembered as \(a = \frac{\Delta v}{t}\), where \(\Delta v\) is the change in velocity (\(v_f - v_i\)) and \(t\) is the time over which this change occurs.\

In the go-cart problem, we used this formula by subtracting the initial velocity from the final velocity and dividing by the time interval to find the go-cart's acceleration. It’s important to keep units consistent, using meters per second squared \(m/s^2\) for acceleration in this case. This step is fundamental because it sets up the ability to apply Newton's second law and solve for the unknown mass.
Mass Calculation in Action
Mass is a measure of the amount of matter in an object and is fundamental to the science of physics, especially when using Newton's second law. In the context of the go-cart exercise, once you've determined the acceleration, you can use the force provided to calculate the mass.

Mass is typically measured in kilograms (kg) in the SI system. The provided exercise uses the formula derived from Newton's second law (\( m = \frac{F}{a} \) ) to find the mass. After substituting the values for force and acceleration into this equation, the result gives you the combined mass of the go-cart and the driver. This process of working backwards from force to find mass is a powerful tool in physics problem-solving.
Physics Problem Solving Techniques
Solving physics problems often involves a step-by-step approach, where we dissect a problem into its fundamental concepts. In our go-cart example, we break down the problem into two main stages: first, finding the acceleration, and second, using it to calculate the mass.

Some key techniques in this approach include:
  • Identifying the known and unknown variables
  • Choosing the correct formulas
  • Substituting the known values and solving for the unknowns
  • Checking the units to ensure they are consistent and make sense
  • Reviewing the answer to see if it's reasonable in the context of the problem
It is this systematic approach that helps ensure accuracy and understanding. By learning and applying these problem-solving strategies, students can become proficient in tackling a wide range of physics problems beyond the textbook exercises.

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Most popular questions from this chapter

The Tornado is a carnival ride that consists of vertical cylinder that rotates rapidly aboat its vertical axis. As the Tornado rotates, the riders are presscd against the inside wall of the cylinder by the rotation, and the floon of the cylinder drops away. The force that points upward. preventing the riders from falling downward, is a) friction force c) gravity. b) a normal force. d) a tension force.

A bosun's chair is a device used by a boatswain to lift himself to the top of the mainsail of a ship. A simplified device consists of a chair, a rope of negligible mass, and a frictionless pulley attached to the top of the mainsail. The rope goes over the pulley, with one end attached to the chair, and the boatswain pulls on the other end, lifting himself upward. The chair and boatswain have a total mass \(M=90.0 \mathrm{~kg}\). a) If the boatswain is pulling himself up at a constant speed, with what magnitude of force must he pull on the rope? b) If, instead, the boatswain moves in a jerky fashion, accelerating upward with a maximum acceleration of magnitude \(a=2.0 \mathrm{~m} / \mathrm{s}^{2},\) with what maximum magnitude of force must he null on the rone?

A block of mass \(M_{1}=0.640 \mathrm{~kg}\) is initially at rest on a cart of mass \(M_{2}=0.320 \mathrm{~kg}\) with the cart initially at rest on a level air track. The coefficient of static friction between the block and the cart is \(\mu_{s}=0.620\), but there is essentially no friction between the air track and the cart. The cart is accelerated by a force of magnitude \(F\) parallel to the air track. Find the maximum value of \(F\) that allows the block to accelerate with the cart, without sliding on top of the cart.

A skier starts with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\) and skis straight down a slope with an angle of \(15.0^{\circ}\) relative to the horizontal. The coefficient of kinetic friction between her skis and the snow is \(0.100 .\) What is her speed after 10.0 s?

A hanging mass, \(M_{1}=0.50 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to the front of a mass \(M_{2}=1.50 \mathrm{~kg}\) that is initially at rest on a frictionless table. A third mass \(M_{3}=2.50 \mathrm{~kg}\), which is also initially at rest on a frictionless table, is attached to the back of \(M_{2}\) by a light string. a) Find the magnitude of the acceleration, \(a,\) of mass \(M_{3}\) b) Find the tension in the string between masses \(M_{1}\) and \(M_{2}\).

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