Question

find the kinetic energy is displacement s=2tsquare +4t at 2secound body mass if of 100g​

Answer 1

Given :

• Mass of the body,m=100g=0.1 kg
• Displacement of body , $\sf\:s=2t^2+4t$

To Find :

Kinetic energy of the body at t = 2 sec

Theory

• Velocity

The rate of change of displacement of a particle with time is called velocity of the particle.

${\purple{\boxed{\large{\bold{Velocity=\frac{Displacement}{Time\:interval}}}}}}$

In differential form:

$\sf\:Velocity,V=\dfrac{dx}{dt}$

• Kinetic energy

It is the engery possessed by a body by virtue of its motion.

Kinetic energy = Work done to put the body into motion - Work done to bring the body at rest

Solution :

We have ,

$\rm\:s=2t^2+4t$

Now , Differentiate with respect to t

$\sf\implies\dfrac{ds}{dt}=\dfrac{d(2t^2)}{dt}+\dfrac{d(4t)}{dt}$

$\sf\implies\dfrac{ds}{dt}=4t+4$

$\sf\implies\:v=4t+4$

At t = 2 sec

$\sf\implies\:v=4(2)+4$

$\sf\implies\:v=8+4$

$\sf\implies\:v=12ms^{-1}$

We know that :

$\sf\blue{Kinetic\:\:energy=\dfrac{1}{2}mv^2}$

Now , Put the given values ,then

$\sf\implies\:K.E=\dfrac{1}{2}\times(0.1)\times(12)$

$\sf\implies\:K.E=\dfrac{1}{2}\times\dfrac{1}{10}\times12$

$\sf\implies\:K.E=\dfrac{6}{10}$

$\sf\implies\:K.E=0.6J$

Therefore , the Kinetic energy of the body is 0.6J

______________________

More information about topic

1. Velocity is a vector quantity
2. The velocity of an object can be positive, zero and negative.
3. SI unit of Velocity is m/s
4. Dimension of Velocity: $\sf\:[M^0LT{}^{-1}]$
5. Kinetic energy can never be negative.
6. Kinetic energy depend on the frame of reference.

Answer 2

Answer:

Given :

Mass of the body,m=100g=0.1 kg

Displacement of body , \sf\:s=2t^2+4ts=2t

2

+4t

To Find :

Kinetic energy of the body at t = 2 sec

Theory

• Velocity

The rate of change of displacement of a particle with time is called velocity of the particle.

{\purple{\boxed{\large{\bold{Velocity=\frac{Displacement}{Time\:interval}}}}}}

Velocity=

Timeinterval

Displacement

In differential form:

\sf\:Velocity,V=\dfrac{dx}{dt}Velocity,V=

dt

dx

• Kinetic energy

It is the engery possessed by a body by virtue of its motion.

Kinetic energy = Work done to put the body into motion - Work done to bring the body at rest

Solution :

We have ,

\rm\:s=2t^2+4ts=2t

2

+4t

Now , Differentiate with respect to t

\sf\implies\dfrac{ds}{dt}=\dfrac{d(2t^2)}{dt}+\dfrac{d(4t)}{dt}⟹

dt

ds

=

dt

d(2t

2

)

+

dt

d(4t)

\sf\implies\dfrac{ds}{dt}=4t+4⟹

dt

ds

=4t+4

\sf\implies\:v=4t+4⟹v=4t+4

At t = 2 sec

\sf\implies\:v=4(2)+4⟹v=4(2)+4

\sf\implies\:v=8+4⟹v=8+4

\sf\implies\:v=12ms^{-1}⟹v=12ms

−1

We know that :

\sf\blue{Kinetic\:\:energy=\dfrac{1}{2}mv^2}Kineticenergy=

2

1

mv

2

Now , Put the given values ,then

\sf\implies\:K.E=\dfrac{1}{2}\times(0.1)\times(12)⟹K.E=

2

1

×(0.1)×(12)

\sf\implies\:K.E=\dfrac{1}{2}\times\dfrac{1}{10}\times12⟹K.E=

2

1

×

10

1

×12

\sf\implies\:K.E=\dfrac{6}{10}⟹K.E=

10

6

\sf\implies\:K.E=0.6J⟹K.E=0.6J

Therefore , the Kinetic energy of the body is 0.6J

______________________

More information about topic

Velocity is a vector quantity

The velocity of an object can be positive, zero and negative.

SI unit of Velocity is m/s

Dimension of Velocity: \sf\:[M^0LT{}^{-1}][M

0

LT

−1

]

Kinetic energy can never be negative.

Kinetic energy depend on the frame of reference.