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oksian1 [2.3K]
3 years ago
12

A car drives around a horizontal, circular track at constant speed. Consider the following three forces that act on the car: (1)

The upward normal force exerted on the car by the road, (2) the downward gravitational force on the car, (3) and the frictional force that is directed toward the center of the circular path.
Which of these forces does zero work on the car as the car moves along the circular path?  1, 2, and 3   3   1 and 2 
 1   2   1 and 3 
 2 and 3 
Physics
1 answer:
musickatia [10]3 years ago
6 0

Answer: 1, 2 and 3.

Explanation:

By definition, work, is the process through which a force, applied on an object, produces a displacement of this object, and can be expressed as the dot product of the force vector and the displacement vector.

It can be also understood as the projection of the force in the direction of the displacement, so, if both vectors are perpendicular, as the projection of one vector along the other is nul, if a force and the displacement are perpendicular, no work is done.

In our case, the 3 forces mentioned, are perpendicular respect the displacement.

For normal force and gravity, as they act vertically, and the car moves along an horizontal trajectory, both are perpendicular, so no work is done.

For the friction force, as it is the only horizontal force present, it is just the centripetal force that keeps the car moving in a circular path.

It always aims to the center of the circle, along the radius at any point, so it is always perpendicular to the displacement also.

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Answer:

1.8 cm

Explanation:

m = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

q = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

\Delta V =Potential difference through which the ion is accelerated = 215 V

v = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}

r = Radius of the path followed by ion

B = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm

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