Question

An automobile has a mass of 1000 kg what
centripetal force is necessary for this vehicle to
make a uniform 90° turn at 72 km/hr in a distance
of 1570 m along the road?
(A) 200 N
(B) 400 N
(C) 600 N
(D) 800 N​

Answer 1

Given:

An automobile has a mass of 1000 kg. This vehicle to make a uniform 90° turn at 72 km/hr in a distance of 1570 m.

To find:

Centripetal force?

Calculation:

• 72 km/hr = 72 × 5/18 = 20 m/s.

First let's calculate the radius :

$\therefore \dfrac{2\pi r}{4} = 1570$

$\implies \dfrac{\pi r}{2} = 1570$

$\implies \dfrac{(3.14)r}{2} = 1570$

$\implies 1.57 \: r= 1570$

$\implies r= 1000 \: m$

Now, centripetal force:

$F = \dfrac{m {v}^{2} }{r}$

$\implies F = \dfrac{1000 \times {(20)}^{2} }{100}$

$\implies F = 400 \: N$

So , centripetal force is 400 N.

Answer 2

Answer:

$Given:</p><p></p><p>An automobile has a mass of 1000 kg. This vehicle to make a uniform 90° turn at 72 km/hr in a distance of 1570 m.</p><p></p><p>To find:</p><p></p><p>Centripetal force?</p><p></p><p>Calculation:</p><p></p><p>72 km/hr = 72 × 5/18 = 20 m/s.</p><p></p><p>First let's calculate the radius :</p><p></p><p>\therefore \dfrac{2\pi r}{4} = 1570∴42πr=1570</p><p></p><p>\implies \dfrac{\pi r}{2} = 1570⟹2πr=1570</p><p></p><p>\implies \dfrac{(3.14)r}{2} = 1570⟹2(3.14)r=1570</p><p></p><p>\implies 1.57 \: r= 1570⟹1.57r=1570</p><p></p><p>\implies r= 1000 \: m⟹r=1000m</p><p></p><p>Now, centripetal force:</p><p></p><p>F = \dfrac{m {v}^{2} }{r}F=rmv2</p><p></p><p>\implies F = \dfrac{1000 \times {(20)}^{2} }{100}⟹F=1001000×(20)2</p><p></p><p>\implies F = 400 \: N⟹F=400N</p><p></p><p>So , centripetal force is 400 N.</p><p></p><p>$

solution by rohit