Question:
$300\, J$ of work is done in sliding a $2 \,kg$ block up an inclined plane of height $10 \,m$. The work done against friction is ( Take $g=10\, m / s ^{2}$ )
$300\, J$ of work is done in sliding a $2 \,kg$ block up an inclined plane of height $10 \,m$. The work done against friction is ( Take $g=10\, m / s ^{2}$ )
Updated On: Jun 14, 2022
- zero
- 100 J
- 200 J
- 300 J
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The Correct Option is B
Solution and Explanation
Work done against friction is equal to
$W' = W - U$
where $U$ is potential energy, $W$ the work done in sliding the block up the inclined plane.
$U = mgh $
$=2\times 10\times 10=200\,J $
$W = 300\, J$
$\therefore W' = 300 - 200 = 100\, J$
$W' = W - U$
where $U$ is potential energy, $W$ the work done in sliding the block up the inclined plane.
$U = mgh $
$=2\times 10\times 10=200\,J $
$W = 300\, J$
$\therefore W' = 300 - 200 = 100\, J$
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Concepts Used:
Work-Energy Theorem
The work and kinetic energy principle (also known as the work-energy theorem) asserts that the work done by all forces acting on a particle equals the change in the particle's kinetic energy. By defining the work of the torque and rotational kinetic energy, this definition can be extended to rigid bodies.
The change in kinetic energy KE is equal to the work W done by the net force on a particle is given by,
W = ΔKE = ½ mv2f − ½ mv2i
Where,
vi → Speeds of the particle before the application of force
vf → Speeds of the particle after the application of force
m → Particle’s mass
Note: Energy and Momentum are related by, E = p2 / 2m.