# How to calculate the Total Work Done?

• Last Updated : 26 May, 2022

When we apply force ‘F’ to a block, the body accelerates or speeds up or down depending on the direction of the force. The kinetic energy of the system changes as the speed increases or decreases. We know that energy cannot be created or destroyed, thus it must be changed into another form. It is referred to as completed work in this situation. When negative energy is done, the energy declines, and when positive work is accomplished, the energy increases. Now we’ll see how to evaluate completed work.

### What is Work Done?

If and only if a force is exerted on a body and the body is moved to a particular displacement as a result of the exerted force, the action is called “work done.” It is denoted by “W”. It is measured in Joules(J).

### How to Calculate Work Done?

When the point of application of a force moves along the force’s path of action, work is completed. The force’s line of action is a line drawn in the force’s direction from the place of application. Assume a constant force vector F acts on an object, causing it to move through a displacement vector s in a direction parallel to the force’s line of action.

So, the Work Done W will be,

W = F.s

or

W = Fscosθ

Work completed is a scalar quantity. It uses SI joule units (J). When the point of application of the force is moved by 1 m and the force has a component of 1 N in the displacement direction, 1 joule of work is done.

If the force is a function of position x rather than a constant, the work done by the force to move the object from position x 1 to position x 2 is given by, If a force vs. distance graph is produced, the labor required to move an object from x1 to x2 is equal to the area beneath the graph between x=x1 and x=x2.

### Sample Problems

Problem 1: A child in a toy cart being pulled ahead by a buddy at a playground, who pushes the cart forward with a force of 60 N along a rope linked to the cart. The rope forms a 35° angle with the ground. Calculate the amount of work done by the child’s playmate to propel the child 20 meters forward.

Solution:

Given,

Force F = 60 N

θ = 35°

s = 20 m

Using Work Done formula,

W = Fscosθ

= (60)(20)(cos35°)

= 980

So, the Work Done is 980 J.

Problem 2: A 15-meter displacement is produced by pulling a box with a force of 25 N. Find the work done by the force if the angle between the force and the displacement is 30°.

Solution:

Given,

Force F = 25N

s = 15 m

θ = 30°

Using Work Done Formula,

W = Fscosθ

= (25)(15)(cos30°)

= 324.76

So, the Work Done is 324,76 J.

Problem 3: Consider a wooden box being dragged across the floor by a wire at a 30° angle from the horizontal floor. The wooden box is dragged 20 meters with a 220 N force applied by the rope. What is the force’s final work product?

Solution:

Given,

Force F = 220 N

The angle between floor and force is 60°

θ = 60°

s =20 m

Using Work Done Formula,

W = Fscosθ

= (220)(20)(cos60°)

= 2200

So, the work Done is 2200 J.

Problem 4: Determine the work done when a force of magnitude 24 N acts at a distance of 10 m in the direction of the force.

Solution:

Given,

F = 24 N

S = 10 m

Using Work Done formula,

W = Fscosθ

= 24×10

= 240

So, the Work Done is 240 J.

Problem 5: When is work done termed zero?

Solution:

The Work Done is zero in the following conditions,

• When the displacement is zero no matter how much force is applied.
• Although the object is moving, the force exerted is zero.
• When the force applied and the object’s movement are in opposite directions.
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