Question

a particle of mass 80 g is moving with a speed of 40ms due to some resistance its velocity is reduced to 10ms find the work done by the resistance​

Answer 1

Question :

A particle of mass 80 g is moving with a speed of 40ms due to some resistance its velocity is reduced to 10ms find the work done by the resistance.

Solution :

From the Question,

Mass of the particle,M = 80 g = 8 × 10^-2 Kg

Initial Velocity,u = 40 m/s

Final Velocity,v = 10 m/s

To finD

Work Done by the air resistance

Work Energy Theorem

$\boxed{ \boxed{ \tt{W = \dfrac{M}{2}( {v}^{2} - {u}^{2}) }}}$

The Work Done by all the forces is equal to the change in Kinetic Energy

Substituting the values,we get :

$\colon \: \implies \: \tt{W = \dfrac{8 }{2} \times {10}^{- 2} ( {40}^{2} - {10}^{2} )} \\ \\ \colon \: \implies \tt{W = \dfrac{4}{100}(1600 - 100) } \\ \\ \colon \: \implies \: \tt{W = 4 \times \frac{1}{100} \times 1500} \\ \\ \colon \: \implies \boxed{\boxed{\tt{W = 60 \: J}}}$

Answer 2

Answer:

$\large\boxed{\sf{60\:J}}$

Explanation:

Mass of particle, m = 80 g = 0.08 kg

Initial velocity, u = 40 m/s

Final velocity, v = 10 m/s

To find the Work Done:

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

And, also we know that,

Kinetic energy, $K.E = \frac{1}{2} m {v}^{2}$

Therefore, we have,

$= > \triangle K.E = \frac{1}{2} m( {u}^{2} - {v}^{2} ) \\ \\ = > \triangle K.E = \frac{1}{2} \times 0.08 \times \left[{(40)}^{2} - {(10)}^{2} \right] \\ \\ = > \triangle K.E = 0.04 \times (1600 - 100) \\ \\ = > \triangle K.E = 0.04 \times 1500 \\ \\ = > \triangle K.E = 60$

Therefore, change in kinetic energy = 60 J

Hence, Work done = 60 J