# Simple work and accleration problem

• highcontrast
In summary, the crate accelerates from 1m/s to 3m/s in 2 seconds under the influence of a 250N force and the floor's friction.
highcontrast

## Homework Statement

A rope exerts a constant horiztonal force of 250N to pull a 60kg crate across the floor. The velocity of the crate is observed to increase from 1m/s to 3m/s in a time of 2 seconds under the influence of this force and the frictional force exerted by the floor on the crate
A) what is the accleration of the crate?
b) what is the netforce acting on the crate?
c) what is the magnitude of the frictional force acting on the crate?

## Homework Equations

As stated in the attempt

## The Attempt at a Solution

a) because of the given information i use the difference in velocity by time to produce a result of 1m/s. I am not sure if i should be a = fnet/mass

b) i use the eq force = ma, 60kg * 1m/s = 60N

c) This is where i am having difficulty, i know for the accleration to be 1m/s, the equation would ultimatey have to be a = fnet/mass = 60n/60kg. subtracting 60n from the 250 yeilds the number, 190, which is the one i need to subtract from the given force to get 60. but i am having a hard time understanding how a 60N force on a crate can produce 130n in friction? Am i way off with my attempts?

Welcome to PF!

Hi highcontrast! Welcome to PF!

(your a and b are fine, and no you don't need to use the mass to get the acceleration from the velocities … it's just a matter of geometry )
highcontrast said:
c) This is where i am having difficulty, i know for the accleration to be 1m/s, the equation would ultimatey have to be a = fnet/mass = 60n/60kg. subtracting 60n from the 250 yeilds the number, 190, which is the one i need to subtract from the given force to get 60. but i am having a hard time understanding how a 60N force on a crate can produce 130n in friction? Am i way off with my attempts?

why are you subtracting 60 twice?

applied force 250 one way, friction force 190 the other way, so net force = 60, so acceleration = 60/60 = 1 …

That's a much more simple way of puting it! I think I may just be confused about the concept of 'net force' and not all that confident in my abilities. The answer definitely made sense in my head. The notion of 60N force acting on the box is what perplexed me. I, somehow, assumed that to be the netforce acting only on the box, but what I failed to realize is that its one system, meaning that 60N netforce would have been the result of the frictional force, and the pulling force.. Am I right in this line of thought?

Hi highcontrast!

Yes, the net force (the total force) is all the external forces on the body,

and it's always the F in F = ma.

As you say, the box is "one system" …

so in the end we can always lump all the forces together as "one force" (and similarly "one torque").

## 1. What is meant by "simple work and acceleration problem"?

Simple work and acceleration problem refers to a type of physics problem that involves calculating the amount of work done on an object by a force and the resulting acceleration of the object. This type of problem typically involves using the equation W = Fd to solve for work and Newton's second law (F = ma) to solve for acceleration.

## 2. How do you determine the work done on an object?

The work done on an object can be determined by multiplying the force applied to the object by the distance the object moves in the direction of the force. This can be represented by the equation W = Fd, where W is work, F is force, and d is distance.

## 3. What is the relationship between work and acceleration in a simple work and acceleration problem?

In a simple work and acceleration problem, the work done on an object is directly proportional to the resulting acceleration of the object. This means that as the amount of work done on the object increases, the acceleration of the object also increases.

## 4. Can you provide an example of a simple work and acceleration problem?

One example of a simple work and acceleration problem is a car accelerating from rest to a certain speed. In this scenario, the work done on the car by the engine is equal to the force of the engine multiplied by the distance the car travels. This work results in an acceleration of the car, which can be calculated using Newton's second law.

## 5. What are some real-world applications of simple work and acceleration problems?

Simple work and acceleration problems have many real-world applications, such as calculating the amount of force needed to lift an object with a pulley system, determining the acceleration of a rocket during launch, and figuring out the amount of work done by a person pushing a box across the floor. These types of problems are used in various fields, including engineering, physics, and mechanics.

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