Question

The speed of the vehicle of mass 500kg increases from 36 km per hour to 72 km per hour calculate the increase in its kinetic energy

answer please ​

Answer 1

Answer:

7.5x10⁴J answer is

Explanation:

Ke=7.5x10⁴J

Answer 2

ProvidEd that:

• Mass of vehicle = 500 kg
• Initial velocity = 36 kmph
• Final velocity = 72 kmph

To calculaTe:

• Increase in kinetic energy

SoluTion:

• Increase in kinetic energy = 75000 Joules or 75 kJ or 7.5 × (10)⁴ Joules.

Using concepts:

• Formula to find initial kinetic energy

• Formula to find final kinetic energy

• Formula to find change in kinetic energy

• Formula to convert kmph-mps

Using formulas:

• Formula to find initial kinetic energy,

• ${\small{\underline{\boxed{\pmb{\sf{K.E_{(initial)} \: = \dfrac{1}{2} \: mu^2}}}}}}$

• Formula to find final kinetic energy,

• ${\small{\underline{\boxed{\pmb{\sf{K.E_{(final)} \: = \dfrac{1}{2} \: mv^2}}}}}}$

• Formula to find change in kinetic energy,

• ${\small{\underline{\boxed{\pmb{\sf{K.E_{(final)} \: - K.E_{(initial)}}}}}}}$

• Formula to convert kmph-mps,

• ${\small{\underline{\boxed{\pmb{\sf{1 \: kmph \: = \dfrac{5}{18} \: mps}}}}}}$

Required solution:

~ Firstly let us convert kmph-mps!

Converting 36 kmph-mps!

$:\implies \sf 36 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{36} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 2 \times 5 \\ \\ :\implies \sf 10 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!!!}}}$

Converting 72 kmph-mps!

$:\implies \sf 72 \times \dfrac{5}{18} \\ \\ :\implies \sf \cancel{72} \times \dfrac{5}{\cancel{{18}}} \\ \\ :\implies \sf 4 \times 5 \\ \\ :\implies \sf 20 \: mps \\ \\ {\pmb{\sf{Henceforth, \: converted!!!}}}$

Therefore,

• Initial velocity = 10 mps
• Final velocity = 20 mps

~ Now let's find out the initial kinetic energy by using suitable formula!

$:\implies \sf K.E_{(initial)} \: = \dfrac{1}{2} \: mu^2 \\ \\ :\implies \sf K.E_{(initial)} \: = \dfrac{1}{2} \times 500 \times 10 \times 10 \\ \\ :\implies \sf K.E_{(initial)} \: = \dfrac{1}{2} \times 500 \times 100 \\ \\ :\implies \sf K.E_{(initial)} \: = \dfrac{1}{2} \times 50000 \\ \\ :\implies \sf K.E_{(initial)} \: = 25000 \: Joules$

~ Now let's find out the final kinetic energy by using suitable formula!

$:\implies \sf K.E_{(final)} \: = \dfrac{1}{2} \: mv^2 \\ \\ :\implies \sf K.E_{(final)} \: = \dfrac{1}{2} \times 500 \times 20 \times 20 \\ \\ :\implies \sf K.E_{(final)} \: = \dfrac{1}{2} \times 500 \times 400 \\ \\ :\implies \sf K.E_{(final)} \: = \dfrac{1}{2} \times 200000 \\ \\ :\implies \sf K.E_{(final)} \: = 100000 \: Joules$

Therefore,

• Initial K.E = 25000 J
• Final K.E = 100000 J

~ Now let's find out the increase in its kinetic energy by using suitable formula!

$:\implies \sf K.E_{(final)} \: - K.E_{(initial)} \\ \\ :\implies \sf 100000 - 25000 \\ \\ :\implies \sf 75000 \: J$

Henceforth, the increase in kinetic energy = 75000 Joules or 75 kJ or 7.5 × (10)⁴ Joules.