# Relationship between work and kinetic energy for a horizontal force displacement

### Work (physics) - Wikipedia

Work, Energy, and Power - Lesson 1 - Basic Terminology and Concepts of Work Done by Forces; Potential Energy · Kinetic Energy · Mechanical Energy · Power Those three quantities are force, displacement and the angle between the . of stairs (a height of m) and then pushes it with a horizontal force of N at. The work done by the force F on the object as it undergoes a displacement d is defined as force perpendicular to displacement exerting a N force at angle of 15deg. above the horizontal. If an object's kinetic energy or gravitational potential energy changes, then work is If the man pushes the rock in the direction of the force, he has done work.

These four forces are shown schematically in Figure 7. Since the velocity of the mass is constant, its acceleration is equal to zero. The x and y-components of the net force acting on the mass are given by Since the net force acting on the mass must be zero, the last equation can be used to determine the normal force N: The kinetic friction force fk is given by However, since the net component of the force along the x-axis must also be zero, the kinetic friction force fk is also related to the applies force in the following manner Combining these last two expressions we can determine the coefficient of kinetic friction: The work done by the rope on the mass m can be calculated rather easily: The work done by the friction force is given by The work done by the normal force N and the weight W is zero since the force and displacement are perpendicular.

### Energy and Work

The total work done on the mass is therefore given by This is not unexpected since the net force acting on the mass is zero. However, in many cases this is not a correct assumption. By reducing the size of the displacement for example by reducing the time interval we can obtain an interval over which the force is almost constant. The work done over this small interval dW can be calculated The total work done by the force F is the sum of all dW Example: The Spring An example of a varying force is the force exerted by a spring that is stretched or compressed.

Suppose we define our coordinate system such that its origin coincides with the end point of a spring in its relaxed state see Figure 7. The force exerted by the spring will attempt to return the spring to its relaxed state: Relaxed, Stretched and Compressed Springs.

The larger the spring constant, the stiffer the spring. The work done by the spring on an object attached to its end can be calculated if we know the initial position xi and final position xf of the object: Pendulum in x-y plane 7.

Work in 2D Consider the pendulum shown in Figure 7.

## Work, force and distance

The pendulum is moved from position 1 to position 2 by a constant force F, pointing in the horizontal direction see Figure 7. The mass of the pendulum is m. What is the work done by the sum of the applied force and the gravitational force to move the pendulum from position 1 to position 2?

Method 1 - Difficult The vector sum of the applied force and the gravitational force is shown in Figure 7. The angle between the applied force F and the vector sum Ft is a. Vector sum Ft of Fg and F. In order to calculate the work done by the total force on the pendulum, we need to know the angle between the total force and the direction of motion.

- Work and the work-energy principle
- Calculating the Amount of Work Done by Forces

The distance dr is a function of d[theta]: For a very small distance dr, the angle between dr and Ft will not change. The maximum angle can be easily expressed in terms of r and h: Angle between sum force and direction. The total work done is Using one of the trigonometric identities Appendix, page A15 we can rewrite this expression as Using the equations shown above for Ft cos aFt sin ar cos [theta]max and r sin [theta]max we can rewrite this expression and obtain for W: Method 2 - Easy The total work done on the pendulum by the applied force F and the gravitational force Fg could have been obtained much easier if the following relation had been used: The total work W is the sum of the work done by the applied force F and the work done by the gravitational force Fg.

These two quantities can be calculated easily: And the total work is which is identical to the result obtained using method 1.

**Kinetic Energy, Gravitational & Elastic Potential Energy, Work, Power, Physics - Basic Introduction**

Kinetic Energy The observation that an object is moving with a certain velocity indicates that at some time in the past work must have been done on it. Suppose our object has mass m and is moving with velocity v.

## Work (physics)

Its current velocity is the result of a force F. For a given force F we can obtain the acceleration of our object: Therefore the time at which the mass reaches a velocity v can be calculated: If at that time the force is turned off, the mass will keep moving with a constant velocity equal to v.

In order to calculate the work done by the force F on the mass, we need to know the total distance over which this force acted. This distance d can be found easily from the equations of motion: The work done by the force F on the mass is given by The work is independent of the strength of the force F and depends only on the mass of the object and its velocity. Since this work is related to the motion of the object, it is called its kinetic energy K: Alternative Derivation Consider a particle with mass m moving along the x-axis and acted on by a net force F x that points along the x-axis.

It falls under the influence of gravity through a distance h see Figure 7. What is its velocity at that point? Joules are the same unit that we measure energy in, which makes sense because work is telling you the amount of joules given to or taken away from an object or a system. If the value of the work done comes out to be positive for a particular force, it means that that force is trying to give the object energy.

The work done by a force will be positive if that force or a component of that force points in the same direction as the displacement. And if the value of the work done comes out to be negative, it means that that force is trying to take away energy from the object.

The work done by a force will be negative if that force or a component of that force points in the opposite direction as the displacement.

If a force points in a direction that's perpendicular to the displacement, the work done by that force is 0, which means it's neither giving nor taking away energy from that object. Another way that the work done by a force could be 0 is if the object doesn't move, since the displacement would be 0. So the force you exert by holding a very heavy weight above your head does not do any work on the weight since the weight is not moving. So this formula represents the definition of the work done by a particular force.

But what if we wanted to know the net work or total work done on an object? We could just find the individual amounts of work done by each particular force and add them up.

But there's actually a trick to figuring out the net work done on an object. To keep things simple, let's assume that all the forces already lie along the direction of the displacement. That way we can get rid of the cosine theta term. Since we're talking about the net work done on an object, I'm going to replace F with the net force on that object. Now, we know that the net force is always equal to the mass times the acceleration. So we replace F net with m times a. So we find that the net work is equal to the mass times the acceleration times the displacement.

I want to write this equation in terms of the velocities and not the acceleration times the displacement. So I'm going to ask you recall a 1-D kinematics equation that looked like this. The final velocity squared equals the initial velocity squared plus 2 times the acceleration times the displacement. In order to use this kinematic formula, we've got to assume that the acceleration is constant, which means we're assuming that the net force on this object is constant.