Question

Certain force acting on a 20kg mass changes its velocity from 5m/s to 2m/s.
Calculate the work done by the force.

ㄗㄥモ丹ちモ 丹れち山モ尺​

Answer 1

Given :-

Mass of the object = 20 kg

Final velocity of the object = 2 m/s

Initial velocity of the object = 5 m/s

To Find :-

The work done by the force.

Analysis :-

Here we are given with the mass, final and initial velocity.

Substitute the values accordingly using the formula of kinetic energy expressing the difference between the velocities.

Solution :-

We know that,

• v = Final velocity
• m = Mass
• u = Initial velocity
• w = Work done
• KE = Kinetic energy

Using the formula,

$\underline{\boxed{\sf Kinetic \ energy=\dfrac{1}{2} m(v^2-u^2)}}$

Given that,

Mass (m) = 20 kg

Final velocity (v) = 2 m/s

Initial velocity (u) = 5 m/s

Substituting their values,

⇒ KE = 1/2 × 20 × (2² - 5²)

⇒ KE = 20/2 × (4 - 25)

⇒ KE = 10 × -21

⇒ KE = -210 J

Therefore, the work done by the force is -210 J.

Answer 2

$\Large{\underline{\bf{\color{cyan}GiVeN,}}}$

• Certain force acting on a 20 kg mass & change it's velocity from 5 m/s to 2 m/s.

$\longmapsto\:\:\bf\blue{Mass\:(m)\:=\:20\:kg}$

$\longmapsto\:\:\bf\purple{Initial\:velocity\:(u)\:=\:5\:m/s}$

$\longmapsto\:\:\bf\pink{Final\:velocity\:(v)\:=\:2\:m/s}$

$\Large{\underline{\bf{\color{coral}To\:FiNd,}}}$

• Work done by the force.

$\Large{\underline{\bf{\color{lime}CaLcUlAtIoN,}}}$

$\bf\blue{As\:we\:know\:that,}$

☆ According to Work-energy theorem,

• Work done by the sum of all forces acting on a particle is equal to the change in kinetic energy of the particle.

$\red\bigstar\:\:{\underline{\green{\boxed{\bf{\color{peru}Work\:done\:(W)\:=\:\triangle{K.E}\:}}}}}$

$\bf\pink{Where,}$

• ∆K.E is the change in kinetic energy.

$:\implies\:\:\bf{W\:=\:\dfrac{1}{2}\:m\:(v^2\:-\:u^2)\:}$

$:\implies\:\:\bf{W\:=\:\dfrac{1}{2}\times{20}\times(2^2\:-\:5^2)\:}$

$:\implies\:\:\bf{W\:=\:10\times(4\:-\:25)\:}$

$:\implies\:\:\bf{W\:=\:10\times(-21)\:}$

$:\implies\:\:\bf\purple{W\:=\:-\:210\:Joules}$

[NOTE :- Here negative sign indicates that change in velocity is decreased from initial to final.]

$\Large\bold\therefore$ The work done by the force is -210 J.