Question

a truck of mass 1000 kg increases its speed from 36 km per hour to 72 km per hour. find the increase in its kinetic energy.
pls someone give me the answer​

Answer 1

Energy - Kinetic Energy

Kinetic energy is the energy of motion. Kinetic energy is directly proportional to the mass of the object and to the square of its velocity.

$\boxed{\bf{K.E. = \dfrac{1}{2}mv^2}}$

where, m is the mass of the object and v is the velocity of the object.

As per the question, a truck of mass $1000 \; kg$ increases its speed from $36 \; km^{-1}$ to $72 \; km^{-1}$. So from here we can conclude that initial velocity is $36 \; km^{-1}$ and the final velocity of object is $72 \; km^{-1}$.

In order to find the increase in its kinetic energy, first we have to convert the units of initial and final velocity from km per hour to m per second.

We know that, to convert the unit from km per hour to m per second we have to divide the velocity value by 3.6. [You can also multiply the velocity value by 5/18 to convert the velocity from km per hour to m per second]

Initial velocity, u = $\frac{36}{3.6} = 10 \; ms^{-1}$

Final velocity, v = $\frac{72}{3.6} = 20 \; ms^{-1}$

Now we know that, Kinetic energy is directly proportional to the mass of the object and to the square of its velocity. We have to calculate the  increase in its kinetic energy. Therefore,

$\implies K.E. = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2 \\ \\ \implies K.E. = \dfrac{1}{2}m(v^2 - u^2)$

Now substituting the known values in the above equation/formula, we get:

$\implies K.E. = \dfrac{1}{2} \times 1000 (20^2 - 10^2) \\ \\ \implies K.E. = \dfrac{1}{2} \times 1000 (400 - 100) \\ \\ \implies K.E. = \dfrac{1}{2} \times 1000 \times 300 \\ \\ \implies K.E. = 1 \times 500 \times 300 \\ \\ \implies K.E. = 150000 \\ \\ \implies \boxed{\bf{K.E. = 1.5 \times 10^5}}$

Hence, the increase in its kinetic energy 1.5 × 10⁵ J.

Answer 2

Here, it is given that

We have to find the increase in its kinetic energy,

In order to do that we have to initial and the final velocity units from km per hour to m per second,

As we have,

Initial velocity, u = $36$ km per hour

Final velocity, v = $72$ km per hour

Mass, m = $1000$ kg

Conversion from km per hour to m per second:

Divide the given value of velocity by $3.6$ or we can also multiply given value by $\frac{5}{18}$ for converting the units,

Therefore,

$u=\frac{36}{3.6}=10m/s$

$v=\frac{72}{3.6}=20m/s$

For calculating the increase in its kinetic energy,

$=> K.E=\frac{1}{2} mv^2-\frac{1}{2}mu^2$

Taking m as common it is written as

$=> K.E=\frac{1}{2} m(v^2-u^2 )$

Put the values in the above formula,

$=> K.E=\frac{1}{2}\times 1000[(20)^2-(10)^2 ]$

By simplifying it further we get,

$=> K.E=\frac{1}{2}\times 1000[400-100 ]$

$=> K.E=\frac{1}{2} \times1000[300 ]$

$=> K.E=\frac{1000}{2} \times[300 ]$

$=>K.E=500\times 300$

$=>K.E=1,50,000$

$=>K.E=1.5\times 10^5$

Hence, the increase in its kinetic energy is $1.5\times 10^5$