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# Work Calculator Physics + Online Solver With Free Steps

The **Work Calculator Physics** calculates the value of work done by using the inputs of force and distance entered by the user. Work describes the energy exerted in moving an object with applied force over a certain distance. Moreover, the force must be applied in the direction of the distance.

The calculator does not support the variables as input. You have to enter the **values** of force and distance in metric units. Furthermore, you have to calculate the **effective force** beforehand to make sure the force is in the direction of motion.

## What Is the Work Calculator Physics?

**The Work Calculator Physics is an online tool that uses the product of the two inputs, force and distance (in metric units),**** to find the value of the work done. Additionally, it provides a breakdown of the actual calculation along with the description of the units of work, force, and distance in metric units. **

The **calculator interface** consists of two input boxes separately with the labels “force” and “distance.” All you have to do is enter the desired value of force and distance in the metric units to find the value of work.

## How To Use the Work Calculator Physics?

You can utilize the **Work Calculator Physics** to find the work done on an object by simply entering the values of force and distance in the text boxes.

The guidelines for the calculator’s usage are below.

### Step 1

First of all, we have to make sure that the values of force and distance must be in metric units. That is the value for force must be in **Newtons **and the value for distance must be in **meters. **Moreover, we have to make sure that the force exerted is in the **direction** of the motion **F = f cos$\theta$**

### Step 2

Afterward, enter the desired values of the force and distance in their respective text boxes as labeled on the calculator interface.

### Step 3

Finally, press the **Submit** button to get the results.

### Results

The results of this calculator show up in a pop-up window on the same page and contain 4 sections:

**Input Interpretation:**You will see the values entered as interpreted by the calculator. It lets you verify the correctness of the input and interpretation.**Result:**The value of the work done with its units in “**joules.**”**Basic Dimensions:**Dimensions of the two inputs and products in the form of basic dimensions, namely**mass, length and time**. This conveys the basic units of the work and further verifies the calculator’s results.**Standard Units:**This part shows the metric units used for the inputs i.e., force and distance.

## How Does the Work Calculator Physics Work?

The **Work Calculator Physics** works by taking the product of the input force F and distance s to find the work done W. This is based on the definition of work itself.

### Definition

**Work done on an object is the force exerted on the object to move it in the direction of the force.** The work done W is then the dot product of the inputs F and s.

**W = F.s **

**W = Fs.cos$\theta$ **

Hence, we can see that work is also dependent on the **angle difference** between the force and the displacement covered by the object. Thus, we would always require the **effective force** exerted on the object and multiply it to find the correct work done by the force on the object.

## Solved Examples

### Example 1

An object is moved **15 meters **by a force of **20 N,** exerted in the **same direction** as the distance. Find the value of the **work done** on the object.

### Solution

The input values of force and distance are:

**F = 20N**

**s = 15m**

As we do not need to calculate the effective distance of the object due to the force and distance having a zero angle difference, we can simply solve it by finding the product of the two inputs.

**W = F.s**

**W = 20 . 15**

**W = 300J**

Where W is work done, and J is the unit, joule, for work.

### Example 2

Consider a stationary object. A force of **25 N** is exerted at an angle of **30 degrees** to move it over a distance of **10 meters**. Find the corresponding **work done** on the object.

### Solution

In this example, we have a non-zero value of direction along with distance and force. Hence, for the calculation of work, we would need the effective force applied in the direction of the motion of the object. It can be done as below.

**F = f cos$\theta$ **

**F = 25 cos(30) **

**F = 21.651 N**

Where F is the effective force exerted in the direction of the object’s displacement (approximated to 3 decimal places), and f is the total force exerted on the object.

Now, we calculate the product of F and s to find the Work W.

**W = F.s**

**W = 21.651 . 10**

**W = 216.51 J**

Where W is Work done, and J is the unit, joule, for work.

### Example 3

Suppose that a** 50 N** force is applied to an object “A” and it moves **20 meters** at an angle of **60 degrees **of the force. Find the** work done** on object A.

### Solution

As done previously in example 2, we will first find the effective force applied on object A in the direction of the force by the following steps:

**F = f cos$\theta$**

**F = 50 . cos(60)**

**F = 25 N**

Finally, we find the product of F and s to find the work W.

**W = F.s**

**W = 25 . 20 **

**W = 500 J**

Hence, the work done on object A is 500 J.