Question

a 2 kg body moving on a level surface collides and compresses horizontal springs of force constant k = 2 N/m through 2 m. calculate the velocity of the body while colliding ( the cofficient of kinetic friction between the block and surface is 0.25​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

m = 2 Kg.

S = 2m.

${ \mu = 0.25}$

(Coefficient of Kinetic Friction)

K = 2N/m

$\huge{\bold{\underline{Explanation:-}}}$

By using work - energy theorem.

$\large{\bold{K.E = Potential \: Energy + Work \: Done \: by \: Friction}}$

Applying Formula

$\large{\bold{ \frac{1}{2}mv^2 = \frac{1}{2}Kx^2 + F.s}}$

Now,

(F = ${ \mu}$mg.s)

$\large{ \cancel{ \frac{1}{2}} mv^2 = \cancel{ \frac{1}{2}} Kx^2 + \mu mg.s}$

Substituting the values.

$\large{ \implies 2 \times v^2 = 2 \times (2)^2 + 0.25 \times 2 \times 10 \times 2}$

$\large{ \implies 2 \times v^2 = 8 + 10}$

$\large{ \implies 2 \times v^2 = 18}$

$\large{ \implies v^2 = \frac{ \cancel{18}}{ \cancel{2}}}$

$\large{ \implies v = \sqrt{9}}$

$\huge{\boxed{\boxed{v = 3 \: m/s}}}$

So, the velocity of body when colliding is 3 m/s.