Question

block of mass 10 kg slides on a horizontal road against friction with initial speed 216 km/h. If it stops

finally, then evaluate the work done by friction​​

Answer 1

Step-by-step explanation:

Concept:

♪ Work:

The Scalar product of force and displacement is known as workdone.

w = F.s

w = F. s cos θ

♪ Work - Energy Theorem:

$\implies{\sf{w = \Delta K.E = K_{f}-K_{i}\:\:\:\:..(1)}}$

Now from 3rd equation of motion,

$\implies{\sf{v^{2}=u^{2}+2as}}$

$\implies{\sf{v^{2}-u^{2}=2as}}$

Multiply m/2 both the sides,

$\implies{\sf{\dfrac{mv^{2}}{2}-\dfrac{mu^{2}}{2}=\dfrac{m}{2}(2as)}}$

$\implies{\sf{(K.E.)_{f}-(K.E.)_{i}=F.s}}$

$\implies{\sf{w=(K.E.)_{f}-(K.E.)_{i}}}$

$\implies{\boxed{\sf{w=\Delta K.E}}}$

Question:

Block of mass 10 kg slides on a horizontal road against friction with initial speed 216 km/h. If it stops finally, then evaluate the work done by friction.

Answer:

Work done by friction = -18 KJ

Step By Step Explanation:

Given:

Mass of block (m) = 10 kg

Initial velocity (u) = 216 km/h

Final velocity (v) = 0 m/s

To find:

Work done by friction.

Formula used:

Work energy formula.

Firstly, we will convert initial velocity km/h in m/s.

$\implies{\sf{u=216\:km/h}}$

$\implies{\sf{u=216\times \dfrac{5}{18}\:m/s}}$

$\implies{\sf{u=216\times \dfrac{5}{18}\:m/s}}$

$\implies{\sf{u=12\times 5\:m/s}}$

$\implies{\boxed{\sf{u=60\:m/s}}}$

Now, put the values in the formula.

$\implies{\sf{workdone =\Delta K.E}}$

$\implies{\sf{workdone = K_{f}-K_{i}}}$

$\implies{\sf{workdone = \dfrac{1}{2}mv^{2}-\dfrac{1}{2}mu^{2}}}$

$\implies{\sf{workdone = \dfrac{1}{2}\times 10\times 0^{2}-\dfrac{1}{2}\times 10 \times 60^{2}}}$

$\implies{\sf{workdone = 0 - \dfrac{1}{2}\times 10 \times 3600}}$

$\implies{\sf{workdone= 0 - \dfrac{1}{2}\times 36000}}$

$\implies{\sf{workdone = 0 - 18000}}$

$\implies{\sf{workdone = - 18000\:Joules}}$

Now, we know that 1 KJ = 1000 J

So, - 18000 J = -18 KJ

Hence, Workdone by friction = -18 KJ.

Answer 2

Step-by-step explanation:

Refer to this attachment