Work, Energy and Power Concept Page - 2

Definition

Positive, Negative and Zero work done

1. Positive Work : If a force displaces the object in its direction, then the work done is positive. The example of this kind of work done is motion of ball falling towards ground where displacement of ball is in the direction of force of gravity.

2. Negative Work : If the force and the displacement are in opposite directions, then the work is said to be negative. For example if a ball is thrown in upwards direction, its displacement would be in upwards direction but the force due to earths gravity is in the downward direction.

3. Zero Work : If the directions of force and the displacement are perpendicular to each other, the work done by the force on the object is zero.
For example, when we push hard against a wall, the force we are exerting on the wall does no work, because in this case the displacement of the wall is d = 0. However, in this process, our muscles are using our internal energy and as a result we get tired.

Definition

Mechanical Energy

Mechanical energy is the energy of an object due to its position and motion. It is equal to the sum of kinetic energy and potential energy. Example: A freely falling body is comprised of mechanical energy.

Definition

Units of Energy

Some units of energy are:

1. SI unit: Joule, 1 J=1 kgms−2$1J=1kgm{s}^{-2}$
2. Electrical unit: 1 kWh=3.6×106 J$1kWh=3.6\times {10}^{6}J$
3. Electron energy: 1 eV=1.6×10−19 J$1eV=1.6\times {10}^{-19}J$
4. 1 calorie=4.184 J$1calorie=4.184J$

Definition

Work Done by Gravity

Gravitational force is the force acting on a body due to gravity.
Example: For a freely falling particle, the particle moves in the direction of gravity. Hence work done by gravity, W=mgh$W=mgh$
Note: For a particle falling under gravity, work done by gravity is dependent only on the difference in height and independent of any vertical displacement.

Example

Work done by frictional force

Example: A 5.0 kg$5.0kg$ box rests on a horizontal surface. The coefficient of
kinetic friction between the box and the surface is 0.5$0.5$. A horizontal force pulls the box at a constant velocity for 10 cm$10cm$. What will be the work done by the applied horizontal force and the frictional force ? (take g=10 m/s2$g=10m/s^2$)

Work done by friction force
=∫F⃗ .ds⃗ $=\int \overrightarrow{F}.d\overrightarrow{s}$
=−(μkN)×10−1$=-({\mu }_{k}N)\times {10}^{-1}$
=−0.5×5×10×10−1$=-0.5\times 5\times 10\times {10}^{-1}$
=−2.5J$=-2.5J$
Using work energy theorem
Work done by force + Work done by friction =ΔK.E$=\mathrm{\Delta }K.E$
Work done by force −2.5$-2.5$ =0$=0$
⇒$\Rightarrow$ Work Done by force =2.5J$=2.5J$

Example

Work Done by a Variable Force

Example: A bicycle chain of length 1.6$1.6$ m and of mass 1$1$ kg is lying on a horizontal floor. What is the work done in lifting it with one end touching the floor and the other end 1.6$1.6$ m above the floor? (Take g=10 ms−2$g=10m{s}^{-2}$)
Solution:
Consider chain to be concentrated on its center of mass.
Therefore, change in potential energy = work done by external force.
Work done=mgh2=8$=mg{\displaystyle \frac{h}{2}}=8$ J

Example

Application of work-energy theorem for a body under constant external force

Example:
A 500$500$ gm ball moving at 15$15$ m/s slows down uniformly until it stops. If the ball travels 15$15$ m, what was the average net force applied while it was coming to a stop?
Solution:
Initial kinetic energy, KE1=12mu2=12×0.5×152=56.25J$K{E}_1={\displaystyle \frac{1}{2}}m{u}^2={\displaystyle \frac{1}{2}}\times 0.5\times {15}^2=56.25J$Final kinetic energy, KE2=12mv2=0J$K{E}_2={\displaystyle \frac{1}{2}}m{v}^2=0J$
Change in kinetic energy, ΔKE=−56.25J$\mathrm{\Delta }KE=-56.25J$
By work energy theorem, work done equals change in kinetic energy.
W=ΔKE=−56.25J$W=\mathrm{\Delta }KE=-56.25J$
For a constant force, work done is W=F.s$W=F.s$
F×15=−56.25$F\times 15=-56.25$
F=−3.75N$F=-3.75N$

Example

Application of work-energy theorem in problems involving gravity

Example:
A particle of mass 100$100$ g is thrown vertically upwards with a speed of 5$5$ m/s. Find the work done by the force of gravity during the time the particle goes up.
Solution:
Change in kinetic energy is ΔKE=0−12×0.1×25=−1.25J$\mathrm{\Delta }KE=0-{\displaystyle \frac{1}{2}}\times 0.1\times 25=-1.25J$
Using work-energy theorem, work done is given by, W=−1.25J$W=-1.25J$

Example

Application of work-energy theorem in problems involving springs

Example:
A body of mass 2$2$ kg is connected to a spring. Under influence of a constant external force, maximum elongation in the spring is 0.5$0.5$ m. Now the external force is removed, find the maximum speed of the mass. Given spring constant = 1 N/m$1N/m$.
Solution:
Spring force on the body is given by, F=−kx$F=-kx$
Work done by spring force is given by, W=∫Fdx=12k(x2i−x2f)$W=\int Fdx={\displaystyle \frac{1}{2}}k(x_i^2-x_f^2)$
W=12×1(0.52−x2f)$W={\displaystyle \frac{1}{2}}\times 1({0.5}^2-x_f^2)$
By work-energy theorem, this equals the change in kinetic energy. 
ΔKE=W$\mathrm{\Delta }KE=W$
KEf−0=12(0.52−x2f)$K{E}_f-0={\displaystyle \frac{1}{2}}({0.5}^2-x_f^2)$
For maximum speed, KEf$K{E}_f$ is maximum, i.e. xf=0$x_f=0$
12mv2=0.125${\displaystyle \frac{1}{2}}m{v}^2=0.125$
v=0.353 m/s$v=0.353m/s$

Example

Use of Work Energy Theorem in Body Under Frictional Force

Example:  A 2$2$ kg block slides on a horizontal floor with a speed of 4$4$ m/s. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is 15$15$ N and spring constant is 10,000$10,000$ N/m. What is the spring compression?

Solution:The total kinetic energy possessed by the block goes into the potential energy of the spring and the work done against friction.Let x$x$ be the compression of the spring.Thus 12mv2=frx+12kx2${\displaystyle \frac{1}{2}}m{v}^2=f_{r}x+{\displaystyle \frac{1}{2}}k{x}^2$⟹5000x2+15x−16=0$⟹5000x^2+15x-16=0$⟹x=0.055m=5.5cm

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