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using-work energy principle the work done by friction force=change in kinetic energy you will need to know:

R is the contact force, = mg

m mass of the disk,

u initial velocity,

v final velocity ,

now we can use the work energy principle to say $0.1 \times mg\times 5=0-\frac{1}{2} \times m \times v^{2}$ this can be simplified by cancelling the m out from both sides and getting $5 \times 0.1 \times 9.8=\frac{1}{2} \times v^{2}$ $v^{2}=9.8$

THEN THE ANSWER IS $v=3.13$

We know that the work done by friction of friction $W_f = F_f \times d\ = F_n \times \mu \times d$ on flat surfaces.
Also, the kinetic energy of an object is given by $KE = \frac{1}{2} \times m \times v^{2}$ .

In order to be at rest, the velocity must be zero, therefore work done by friction to slow the disk must be equal to the initial kinetic energy of the disk itself:
$KE= W_f$
$\frac{1}{2} \times m \times v^{2} = m\times g \times \mu \times d$
After cancelling out the mass term on both sides and multiplying both sides by 2:
$v^{2} = 2 \times 9.8 \times 0.1 \times 5$
$v = \sqrt{9.8} = \boxed{3.13}$ which is the final answer.

Here the retarding force is equal to $f_k=-\mu_k N=-\mu_k mg$ . Let $-\mu_k mg=ma$ , thus $-\mu_k g=a=v \dfrac{\mathrm dv}{\mathrm dx}=$ $\Rightarrow -\int_0^{x_i} \mu_k g \mathrm dx= \int_{v_i}^0$ Thus $\mu_k g\cdot (x_i)=\dfrac{(v_i)^2}{2}$ . Substituting the values $x_i=5m$ we obtain $v_i=3.130ms^{-1}$ .

K.E = work done by non conservative force 1/2 mv^2= umgl so. v^2=(2ugl)^1/2 v^2=(2 x 0.1 x 9.8 x5) s.r. on both side v=√9.8 v= 3.1304951685

1/2mv^2=umgx.....KE lost due friction=work done by non conservative force, here x=5m

½ mv^2 = umgl so v = (2ugl)^1/2 v = (2 0.1 9.8*5) v = 3.1 m/s

0.1= ma/mg, a=0.98, u=(2x0.98x5)^0.5, =3.13

ma=0.1mg

a=-9.8m/s2

v2=u2+2as

v=0

a=3.130

Energy from friction is what stops the disk, we we find the work from friction

w=fd

w=MND

w=.1* m * 9.8*5

w=4.9m

Work is equivalent to mechanical energy, which is kinetic

4.9m=.5mv^2

Now we can conveniently cancel m from both sides

4.9=.5*v^2

9.8=v^2

3.13=v

m = mass g = acceleration due to gravity N = normal force u = coefficient of kinetic friction d = distance

Newton's second law: net force = ma

Along the vertical, N = mg

Along the horizontal, -friction = ma -uN = ma -umg = ma --> a = -ug

From kinematic equations:

a = [(Vfinal)^2 - (Vinitial)^2]/2d

Since Vfinal = 0

--> v(initial) = sqrt(2ugd)

= 3.13 m/s

1. v=0, s=5 m v2 = u2  2as 02 = u2  10a Now, the co-efficient of kinetic friction is μ f = μN = μmg = ma a = μg = 0.1 × 9.8 = 0.98 Now, u2  10a = 0 u2  10 × 0.98 = 0 u2 = 9.8 u = 3.13