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Chapter 3: Problem 54

A horizontal cable pulls a 200 -kg cart along a horizontal track. The tension in the cable is \(500 \mathrm{~N}\). Starting from rest, \((a)\) How long will it take the cart to reach a speed of \(8.0 \mathrm{~m} / \mathrm{s}\) ? (b) How far will it have gone?

Short Answer

Expert verified
(a) 3.2 seconds, (b) 12.8 meters.

Step by step solution

01

Understanding the Problem

We need to calculate the time it will take for a 200 kg cart to accelerate from rest to a speed of 8.0 m/s given a force of 500 N. Subsequent to that, we need to determine the distance traveled by the cart during this time.
02

Calculate Acceleration

Using Newton's second law of motion, we find the acceleration of the cart. The formula is: \[ a = \frac{F}{m} \]where \( F = 500 \text{ N} \) and \( m = 200 \text{ kg} \). So we have: \[ a = \frac{500}{200} = 2.5 \text{ m/s}^2 \]
03

Calculate Time to Reach Desired Speed

Using the concept of uniform acceleration, the time \( t \) required to reach a velocity \( v \) from rest can be determined by the formula:\[ v = a \times t \]Solving for time, we get:\[ t = \frac{v}{a} = \frac{8.0 \text{ m/s}}{2.5 \text{ m/s}^2} = 3.2 \text{ seconds} \]
04

Calculate Distance Traveled

We calculate the distance traveled using the formula for distance in uniformly accelerated motion, which is:\[ d = \frac{1}{2} a t^2 \]Substitute \( a = 2.5 \text{ m/s}^2 \) and \( t = 3.2 \text{ s} \) into the equation:\[ d = \frac{1}{2} \times 2.5 \times (3.2)^2 \approx 12.8 \text{ meters} \]

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Key Concepts

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

Uniform Acceleration
Uniform acceleration refers to a scenario where an object's velocity changes at a constant rate over a period of time. This means that the object moves faster or slower with no sudden changes in acceleration.
For the cart pulled by the cable, the acceleration is uniform because the force applied is consistent. We calculated the acceleration at 2.5 m/s虏.
With uniform acceleration, you can easily predict how fast an object is moving and how far it has traveled because everything happens steadily and predictably.
  • Velocity increases linearly over time.
  • Formulas like \(v = a \times t\) and \(d = \frac{1}{2} a t^2\) help us solve for time, velocity, or distance.
This makes it much easier to analyze motion in physics problems like this one.
Forces and Motion
Forces and motion are deeply interconnected. Newton's second law of motion gives us a clearer understanding of how forces result in acceleration.
It states that the acceleration \(a\) of an object is directly proportional to the net force \(F\) acting upon the object and inversely proportional to the object's mass \(m\). Simply expressed as \(a = \frac{F}{m}\).
In our exercise, a 500 N force is pulling a 200 kg cart, leading to an acceleration of 2.5 m/s虏. The cart's motion from rest to a certain speed is driven by this applied force.
  • Higher forces result in greater acceleration if the mass remains constant.
  • Understanding this helps us manipulate variables to achieve desired motion outcomes.
Newton's laws provide the foundation for predicting how objects will move under certain forces.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion. This includes concepts like displacement, velocity, and acceleration.
In this exercise, using kinematic equations helps us find out how long it takes a cart to reach a specific speed and how far it travels.
  • The key equations are \(v = a \times t\) for velocity, and \(d = \frac{1}{2} a t^2\) for distance.
  • These equations assume initial conditions, like starting from rest, which simplifies calculations.
Kinematics allows us to precisely describe the motion itself, leading to a better understanding of how objects move in real-world scenarios.

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Most popular questions from this chapter

Typically, a bullet leaves a standard 45-caliber pistol (5.0-in. barrel) at a speed of \(262 \mathrm{~m} / \mathrm{s}\). If it takes \(1 \mathrm{~ms}\) to traverse the barrel, determine the average acceleration experienced by the \(16.2\) -g bullet within the gun, and then compute the average force exerted on it.

The coefficient of static friction between a box and the flat bed of a truck is 0.60. What is the maximum acceleration the truck can have along level ground if the box is not to slide? The box experiences only one \(x\) -directed force, the friction force. When the box is on the verge of slipping, \(F_{f}=\mu \mathrm{s} F_{W}\), where \(F_{W}\) is the weight of the box. As the truck accelerates, the friction force must cause the box to have the same acceleration as the truck; otherwise, the box will slip. When the box is not slipping, \(\sum F_{x}=m a_{x}\) applied to the box gives \(F_{f}=m a_{x}\). However, if the box is on the verge of slipping, \(F_{f}\) \(=\mu_{s} F_{W}\) so that \(\mu_{s} F_{W}=m a_{x} .\) Because \(F_{W}=m g\), $$ a_{x}=\frac{\mu_{s} m g}{m}=\mu_{s} g=(0.60)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=5.9 \mathrm{~m} / \mathrm{s}^{2} $$ as the maximum acceleration without slipping.

Once ignited, a small rocket motor on a spacecraft exerts a constant force of \(10 \mathrm{~N}\) for \(7.80 \mathrm{~s}\). During the burn, the rocket causes the 100 -kg craft to accelerate uniformly. Determine that acceleration.

Just as her parachute opens, a 60 -kg parachutist is falling at a speed of \(50 \mathrm{~m} / \mathrm{s}\). After \(0.80 \mathrm{~s}\) has passed, the chute is fully open and her speed has dropped to \(12.0 \mathrm{~m} / \mathrm{s} .\) Find the average retarding force exerted upon the chutist during this time if the deceleration is uniform.

A 20 -kg crate hangs at the end of a long rope. Find its acceleration (magnitude and direction) when the tension in the rope is ( a) 250 \(\mathrm{N},(b) 150 \mathrm{~N},(c)\) zero, (d) \(196 \mathrm{~N}\).
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