Question

How much work is done to increase a 20-kg scooter's speed from 10 m/s to 20 m/s

a. 1000j
b. 3000j
c. 4000j
d. 100j

Answer 1

3000 Joule of work is to be done.

Option: b

Step-by-step explanation:

Given, mass of scooter = 20 Kg

Initial speed = 10 m/s, Final speed = 20 m/s, Work done = ΔE

Where, ΔE= change in kinetic energy

Solution:

`\text { Kinetic Energy (initial) }=(1)/(2) * \mathrm{mv}^(2)`

`\text { Kinetic Energy (initial) }=(1)/(2) * 20 \mathrm{Kg} *(10 \mathrm{m} / \mathrm{s})^(2)`

`(1)/(2) * 200 \mathrm{Kg} \cdot \mathrm{m}^(2) / \mathrm{s}^(2)` =
`1000 \mathrm{Kg} \cdot \mathrm{m}^(2) / \mathrm{s}^(2)`

`\text { Kinetic Energy (final) }=(1)/(2) * m v^(2)`

`(1)/(2) * 20 \mathrm{Kg} *(20 \mathrm{m} / \mathrm{s})^(2)`

`(1)/(2) * 20 \mathrm{Kg} * 400 \mathrm{m}^(2) / \mathrm{s}^(2)`

`(1)/(2) * 8000 \mathrm{Kg} \cdot \mathrm{m}^(2) / \mathrm{s}^(2)`

`4000 \mathrm{Kg} \cdot \mathrm{m}^(2) / \mathrm{s}^(2)`

Work done = ΔE = Kinetic Energy (final) - Kinetic Energy (initial)

Work done = ΔE = 4000 J - 1000 J work done = ΔE = 3000J