The radius of curvature of a highway exit is r = 93.5 m. The surface of the exit road is horizontal, not banked. (See figure.)

Physics

1 answer:

trapecia [35]2 years ago

8 0

Static friction keeps the car from skidding off the road and points toward the center of the curve. By Newton's second law, the car experiences

• net vertical force

F [normal] - F [weight] = 0

• net horizontal force

F [friction] = ma = mv²/r

where v is the tangential speed of the car.

It follows that

F [normal] = F [weight] = mg

and when static friction is maximized at the car's maximum speed,

F [friction] = µ F[normal] = 0.402 mg

Solve for v :

0.402 mg = mv²/r ⇒ v = √(0.402 g (93.5 m)) ≈ 19.2 m/s

You might be interested in

What had to be invented before we could discover Saturn and Neptune?

Ann [662]

The discovery of Uranus, Neptune, most asteroids, the moons of
any planet other than Earth, Pluto, and all the other dwarf planets,
required the invention and use of telescopes. Those objects are all
too dim to be seen with a bare naked human eye.

Saturn is bright and plainly visible with your eyes, if you know when
and where to look, and what you're looking at.

4 0

3 years ago

Read 2 more answers

If total (PE + KE) is considered

diamong [38]

The sum of Potential Energy and Kinetic Energy is called Mechanical Energy

3 0

4 years ago

A car slows down uniformly from a speed of 22 m/s to rest in 4.0 seconds. How far did it travel in that time

krek1111 [17]

Answer:

$\boxed{\sf Distance \ travelled = 44 \ m}$

Given:

Initial speed (u) = 22 m/s

Final speed (v) = 0 m/s (Rest)

Time taken (t) = 4 seconds

To Find:

Distance travelled by car (s)

Explanation:

From equation of motion of object moving with uniform acceleration in straight line we have:

$\boxed{ \bold{s = (\frac{v + u}{2} )t}}$

By substituting value of v, u & t in the equation we get:

$\sf \implies s = ( \frac{0 + 22}{2} ) \times 4 \\ \\ \sf \implies s = \frac{22}{2} \times 4 \\ \\ \sf \implies s = 11 \times 4 \\ \\ \sf \implies s = 44 \: m$