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Goryan [66]
3 years ago
7

A 1200kg car traveling at 45m/s slams on its breaks and slides to a stop. If the coefficient of kinetic friction between the car

and the road is 0.85 determine how far the car slides
Physics
1 answer:
gavmur [86]3 years ago
6 0

Given :

A 1200 kg car travelling at 45 m/s slams on its breaks and slides to a stop.

If the coefficient of kinetic friction between the car and the road is 0.85 .

To Find :

How far the car slides.

Solution :

Acceleration is given by :

ma = \mu_kmg\\\\a = \mu_kg\\\\a = 0.85\times 10\ m/s^2\\\\a =  8.5\ m/s^2

Distance covered is :

2as=u^2-v^2\\\\s = \dfrac{u^2-v^2}{2a}\\\\s=\dfrac{45^2-0^2}{2\times 8.5}\\\\s=119.12\ m

Therefore, distance covered is 119.12 m.

Hence, this is the required solution.

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Answer:

w = vR/3

Explanation:

The centre of mass of the loop to bullet system is given by D / 4 from centre of loop, which is equivalent to R / 2 from its centre.

From the principle of conservation of linear momentum , we have

m*v = 2*m* Vcm

Where v = velocity of bullet, Vcm = velocity of wood

Hence, we have  

Vcm = v2

Also, from the conservation of angular momentum about the centre of mass.

M*V*(R/2) = Ic*w - equation (I)

where Ic = moment of inertia and w = angular velocity

Ic for a ring is given by mr^2 + m(r/2)^2

Ic of a bullet is given by m(r/2)^2

Hence, the moment of inertia of the system  is given by the summation of the two moments of inertia Ic(ring) + Ic(bullet) which gives

Ic(system) = 3*m*R^2/2

Substituting back into equation (I), we have

m*v*R^2=3*m*R^2*w/2

Hence, we obtain w =vR/3

w=v3R

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