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Volume of a Cylinder Formula with Solved Examples

Explore more about the volume of a cylinder formula with solved examples.

Everything you need to know about the work formula is provided below. Please proceed to read the whole document carefully to understand the topic completely.

The work recipe is used to register how much work is expected to move anything. The result of the applied power and the removal of the applied power is work. The dab result of two vectors, power and relocation, is work. Work is hence a scalar number. The Joule is the SI unit of work (J).

The equation for Work can be communicated as,

W = F.d

W = (Fcos θ)d

Where,

  • W = Work done
  • F = Magnitude of the power applied
  • d = Magnitude of the relocation toward the power.
  • θ = is the point between the vectors: power and removal

SI unit of work is Joule (J). On the off chance that 1 joule of work is finished, the recipe for sort out comes to be 1 J = 1 N.m

Solved Examples

Example 1:

 8 Newtons of force is applied to a body which displaces it by 4 metres. Calculate the work done by using the formula for work.

Solution:

To find: Work done

Given: Force (F) = 8 N

Displacement (d) = 4 m

Using Formula For WorK,  W = F.d

= (8)(4)

= 32Nm

Example 2:

 A coolie at a railway station carries a bag weighing 10 N through some distance. Calculate the work done by the coolie on the bag by using the formula for work.

Solution:

To find work done by the coolie.

Given: Weight of the bag = 10N

Also, the weight of the bag will be acting in the vertical direction and its motion is in the horizontal direction. So the displacement of the bag in the direction of the force (weight) is zero.

d = 0

Using Formula For Work,

W = F.d

= (10)(0)

= 0 J

Answer: The work done by the coolie on the bag is zero.

Example 3: 

Calculate the amount of work done by the force in moving the object through a distance of 5 m if an object is horizontally dragged across the surface by a 300N force acting parallel to the surface.

Solution:

To find: Work done by the force in moving the object through a distance of 5 m

Given: F = 300 N, d = 5 m

Since F and d are in the same direction,

θ = 0, [θ is the angle of the force to the direction of movement]

W = F × Cos θ × d

= 300× 5× Cos 0

= 1500J [Since Cos 0 = 1]

Answer: The amount of work done by the force in moving the object is 1500 J.

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