Question

A body weighing 1000kg accelerates uniformly from 10m/s to 20m/s. Find the amount of work done during this period

Answer 1

Answer:

Work done by the body is 150 k J.

Explanation:

Using work-energy theorem :

• Change in kinetic energy = Net work done by the body.

We know,

• Kinetic events of a particle( moving ) is given by 1 / 2 ( mv^2 ), where m is its mass and v is its velocity.

Here,

Initial velocity v = 10 m/s

Final velocity u = 20 m/s

= > ∆KE = 1 / 2 m( 20^2 ) - 1 / 2 m( 10^2 ) J

= > ∆KE = 1 / 2 m{ 400 - 100 } J

= > Work done = $\dfrac{1}{2}\times1000\times300$ J

= > Work done = 150000 J

= > Work done = 150 kilo J.

Hence work done by the body is 150 k J.

Answer 2

Answer:

$\large \boxed { \sf{1.5 \times {10}^{5}\:J }}$

Explanation:

It's being given that,

Mass of a body, $\sf{m = 1000 \:kg}$

Initial velocity , $\sf{u=10\:m\:{s}^{-1}}$

Final velocity, $\sf{v=20\:m\:{s}^{-1}}$

To find the work done,

• We know that work done is equal to change in kinetic energy.

Also, we know that,

$\large\boxed{\sf{\pink{K.E = \frac{1}{2}m{v}^{2}}}}$

Putting the respective values, We get,

$\sf{= > \triangle K.E = \frac{1}{2} m( {v}^{2} - {u}^{2} ) }\\ \\ \sf{ = > \triangle K.E = \frac{1}{2} m( {(20)}^{2} - {(10)}^{2} )} \\ \\ \sf{= > \triangle K.E = \frac{1}{2} \times 1000 \times (400 - 100)} \\ \\ \sf{= > \triangle K.E = 500 \times 300} \\ \\ \sf{ = > \triangle K.E = 1.5 \times {10}^{5} } \: joules$

Hence, Work done is equal to $\sf{1.5\times{10}^{5}\:J}$