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1800-102-2727Sheeba and her family own a house located at the hilltop which is several meters above the ground level. They own a car parked at their house garage. One day, Sheeba decides to take the car out for a ride. In order for her to reach the bottom of the hill, she had to drive and descend several meters below. As she descends, Sheeba has changed her position under the influence of the force. She has been displaced. It means some work must have been done on her and the car. We know,$Workdone=Force\times Displacement.$
But the question is, where does this work done come from? You might have heard the fact that work done and energy are related. When she was at the top of the hill, her energy was purely potential energy. As she starts driving down the hill, the car gets in motion, hence the potential energy gradually transforms into kinetic energy.
Similarly, in the case of hydroelectric power plants, electricity is generated due to the transformation of kinetic energy of the falling water. The potential energy of the stored water in the reservoir at higher altitude converts into high kinetic energy as the water flows through.
In this article, let us explore kinetic energy in detail.
Table of contents
Kinetic energy (K) is defined as the energy possessed by an object due to its virtue of motion. Let us consider the following example where a block of mass m is placed on a wedge having inclination with the ground. When the block starts to slide down the wedge, its potential energy gets converted into kinetic energy. If v is the velocity acquired by the block as a result of sliding, then the kinetic energy, K is given as,
$K=\frac{1}{2}m{v}^{2}$
It is a scalar quantity. Its SI unit is Joule(J). Its dimensional formula is $\left[M{L}^{2}{T}^{-2}\right].$
Note:
The erg is the CGS unit of energy.
$1erg={10}^{-7}J.$
Let us consider n particles having masses m1, m2 and m3... and their respective velocities are $\overrightarrow{{v}_{1}},\overrightarrow{{v}_{2}},\overrightarrow{{v}_{3}}...$. and so on. The kinetic energy of the system of particles, K_{system} is the algebraic sum of the kinetic energy of the individual particles.
Mathematically,
We have,
$K=\frac{1}{2}m{v}^{2}=\frac{{\left(mv\right)}^{2}}{2m}$
Now p=mv is the linear momentum of the particle.
Hence,
$K=\frac{{p}^{2}}{2m}$
So, p=2Km is the relation between kinetic energy and momentum.
It states that the work done by a force is numerically equal to the change in kinetic energy of the body on which the force acts. Let dW be the work done by an external force F which pushes the body through a distance dx along the force.
Then dW=F.dx
Now, F=ma denotes the force acting on a mass m and the acceleration of the body is a.
So, $\int dW=\int madx;$ where acceleration $a=v\frac{dv}{dx}$
The initial and final velocity of the body are u and v respectively.
So, $W=\int mv\left(\frac{dv}{dx}\right)dx={\int}_{u}^{v}mvdv=\frac{m{v}^{2}}{2}-\frac{m{u}^{2}}{2}$
Hence, work done on the body is numerically equal to the kinetic energy change produced in the body.
Video explanation
Q. A block of mass M slides on a frictionless surface of an inclined plane, as shown in figure. The angle of the inclined plane changes from 600 to 300 at point B. The block starts from rest at point A. Assume that the collisions between the block and any point on the inclined plane are totally elastic $(g=10m{s}^{-2}).$The speed of the block at point B immediately after it hits the second inclined plane will be
(a)$\sqrt{60}m/s$ (b)$\sqrt{45}m/s$(c)$\sqrt{30}m/s$
(d)$\sqrt{15}m/s$
A. b
Let vbe the velocity at point B and the height covered by the block be h.
Then $tan{60}^{0}=\frac{h}{\sqrt{3}};$
h=3 m.
Applying conservation of energy,
$\frac{1}{2}m{v}^{2}=mgh;$
$v=\sqrt{2gh}=\sqrt{2\times 10\times 3}=\sqrt{60}m/s.$
Speed of the block when it strikes the second incline would be the cosine component of v which acts along BC.
${v}_{1}=vcos({60}^{0}-{30}^{0})=vcos{30}^{0}=\sqrt{60}\times \frac{\sqrt{3}}{2}=\sqrt{45}m/s.$
Q. In the above question, calculate the speed of the block at point C, before it loses contact with the second inclined plane
(a)$\sqrt{120}m/s$ (b)$\sqrt{105}m/s$ (c)$\sqrt{90}m/s$ (d)$\sqrt{75}m/s$
A. b
Altitude lost by the block when it falls from B to C would be,
$h\text{'}=3\sqrt{3}tan{30}^{0}=3m.$
Let vc be the speed of the block at point C before it loses contact with the second incline.
Then, conserving mechanical energy at points C and B,
$\frac{1}{2}m{{v}_{c}}^{2}=\frac{1}{2}m{{v}_{1}}^{2}+mgh\text{'}.$
${v}_{c}=\sqrt{{{v}_{1}}^{2}+2gh\text{'}}=\sqrt{45+2\times 10\times 3}=\sqrt{105}m/s.$
Q. A ball of mass 5 kg falls from rest through a distance of 20 m and attains a velocity of 10 m s-1. Calculate the work done by the air resistance on the ball.
(a)-150 J (b)+150 J (c)-250 J (d)-750 J
A. d
Let work done by gravity be Wmg.
Then ${W}_{mg}=mgh=5\times 10\times 20=1000J;$
Let Wair be the work done by the air resistance on the ball.
${W}_{mg}+{W}_{air}=\frac{1}{2}m{v}^{2};{W}_{air}=\frac{1}{2}\times 5\times {\left(10\right)}^{2}-5\times 10\times 20=-750J.$
Q. If the kinetic energy of the particle is increased by 300 %, find the percentage increase in momentum.
(a)100 % (b)200 % (c)400 % (d)500 %
A. a
Let K indicate the initial kinetic energy of the particle and K' be the final kinetic energy.. Then
$K\text{'}=K+\frac{300\%}{100\%}K=4K$
Let pbe the initial momentum and p'be the final momentum. Then,
$K=\frac{{p}^{2}}{2m};K\text{'}=\frac{{p\text{'}}^{2}}{2m}$
Now
$4K=\frac{{p\text{'}}^{2}}{2m}--\left(i\right)$
$K=\frac{{p}^{2}}{2m}--\left(ii\right)$
Dividing (i) by (ii), we get
$4=\frac{{p\text{'}}^{2}}{{p}^{2}};p\text{'}=2p.$
% change in momentum$\frac{p\text{'}-p}{p}\times 100\%=\frac{2p-p}{p}\times 100\%=100\%.$
Q. Can kinetic energy be stored?A. Yes, it can be stored. For instance, when a block having high KE strikes a spring, its KE is stored as spring potential energy.
Q. Does kinetic energy create heat?A. Yes, a ball moving with high kinetic energy on a rough surface generates heat due to friction.In such cases, some amount of energy
Q. Can kinetic energy be negative?.A. K=12mv2. Mass and square of velocity can never be negative. Hence kinetic energy is never negative.
Q. Which form of energy does not involve kinetic energy?A. Light energy doesn’t involve kinetic energy. It is an electromagnetic radiation having electric and magnetic field energy. It has nothing to do with kinetic energy.