Question:

$300\, J$ of work is done in sliding a $2 kg$ block up an inclined plane of height $10 \,m$. Taking $g=10\, m / s ^{2}$, work done against friction is :

Updated On: Jul 27, 2022
  • 200 J
  • 100 J
  • zero
  • 1000 J
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The Correct Option is B

Solution and Explanation

Net work done in sliding a body up to a height h on inclined plane $=$ Work done against gravitational force $+$ Work done againstfrictional force $\Rightarrow W=W_{g}+W_{f} \,\,\,. ..$ (i) but $W=300\, J$ $ W_{g}=m g h=2 \times 10 \times 10=200\, J$ Putting in E (i), we get $300=200+W_{f}$ $\Rightarrow W_{f}=300-200$ $=100\, J$
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Concepts Used:

Work

Work is the product of the component of the force in the direction of the displacement and the magnitude of this displacement.

Work Formula:

W = Force × Distance

Where,

Work (W) is equal to the force (f) time the distance.

Work Equations:

W = F d Cos θ

Where,

 W = Amount of work, F = Vector of force, D = Magnitude of displacement, and θ = Angle between the vector of force and vector of displacement.

Unit of Work:

The SI unit for the work is the joule (J), and it is defined as the work done by a force of 1 Newton in moving an object for a distance of one unit meter in the direction of the force.

Work formula is used to measure the amount of work done, force, or displacement in any maths or real-life problem. It is written as in Newton meter or Nm.