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kolbaska11 [484]

3 years ago

15

A 77 g Frisbee is thrown from a point 0.99 m above the ground with a speed of 15 m/s. When it has reached a height of 1.3 m, its

speed is 12 m/s. What was the reduction in the mechanical energy of the Frisbee-Earth system because of air drag?

1 answer:

Alex73 [517]3 years ago

4 0

Answer:

The reduction in the mechanical energy of the Frisbee-Earth = 2.885 J

Explanation:

From the law of conservation of energy,

Mechanical Energy = Ek + Ep

Where Ek = kinetic Energy, Ep = potential Energy.

Change in mechanical energy = change in kinetic Energy + change in potential energy.

ΔM = ΔEk + ΔEp ................................. Equation 1.

Reduction in mechanical energy = -ΔM = -(ΔEk + ΔEp)

-ΔM = -[1/2m(v₂²-v₁²) + mg(h₂-h₁)]..................... Equation 2

Where -ΔM = Reduction in mechanical Energy, m = mass of the Frisbee, v₂ = final velocity of the Frisbee, v₁ = Initial velocity of the Frisbee, h₂ = Final height of the Frisbee, h₁ = Initial height of the Frisbee.

Given: m = 77 g = 0.077 kg, v₁ = 15 m/s, v₂ = 12 m/s, h₁ = 0.99 m, h₂ = 1.3 m, g = 9.81 m/s²

Substituting these values into equation 2

-ΔM = -[1/2(0.077)(12²-15²) + 0.077(1.3-0.99)(9.81)]

-ΔM = -[(0.077)(-40.5) + (0.234)]

-ΔM = -[-3.119+0.234]

-ΔM = -[-2.8845]

-ΔM ≈ 2.885 J

Thus the reduction in the mechanical energy of the Frisbee-Earth = 2.885 J

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Consider a cyclotron in which a beam of particles of positive charge q and mass m is moving along a circular path restricted by

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A) $v=\sqrt{\frac{2qV}{m}}$

B) $r=\frac{mv}{qB}$

C) $T=\frac{2\pi m}{qB}$

D) $\omega=\frac{qB}{m}$

E) $r=\frac{\sqrt{2mK}}{qB}$

Explanation:

A)

When the particle is accelerated by a potential difference V, the change (decrease) in electric potential energy of the particle is given by:

$\Delta U = qV$

where

q is the charge of the particle (positive)

On the other hand, the change (increase) in the kinetic energy of the particle is (assuming it starts from rest):

$\Delta K=\frac{1}{2}mv^2$

where

m is the mass of the particle

v is its final speed

According to the law of conservation of energy, the change (decrease) in electric potential energy is equal to the increase in kinetic energy, so:

$qV=\frac{1}{2}mv^2$

And solving for v, we find the speed v at which the particle enters the cyclotron:

$v=\sqrt{\frac{2qV}{m}}$

B)

When the particle enters the region of magnetic field in the cyclotron, the magnetic force acting on the particle (acting perpendicular to the motion of the particle) is

$F=qvB$

where B is the strength of the magnetic field.

This force acts as centripetal force, so we can write:

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

where r is the radius of the orbit.

Since the two forces are equal, we can equate them:

$qvB=m\frac{v^2}{r}$

And solving for r, we find the radius of the orbit:

$r=\frac{mv}{qB}$ (1)

C)

The period of revolution of a particle in circular motion is the time taken by the particle to complete one revolution.

It can be calculated as the ratio between the length of the circumference ( $2\pi r$ ) and the velocity of the particle (v):

$T=\frac{2\pi r}{v}$ (2)

From eq.(1), we can rewrite the velocity of the particle as

$v=\frac{qBr}{m}$

Substituting into(2), we can rewrite the period of revolution of the particle as:

$T=\frac{2\pi r}{(\frac{qBr}{m})}=\frac{2\pi m}{qB}$

And we see that this period is indepedent on the velocity.

D)

The angular frequency of a particle in circular motion is related to the period by the formula

$\omega=\frac{2\pi}{T}$ (3)

where T is the period.

The period has been found in part C:

$T=\frac{2\pi m}{qB}$

Therefore, substituting into (3), we find an expression for the angular frequency of motion:

$\omega=\frac{2\pi}{(\frac{2\pi m}{qB})}=\frac{qB}{m}$

And we see that also the angular frequency does not depend on the velocity.

E)

For this part, we use again the relationship found in part B:

$v=\frac{qBr}{m}$

which can be rewritten as

$r=\frac{mv}{qB}$ (4)

The kinetic energy of the particle is written as

$K=\frac{1}{2}mv^2$

So, from this we can find another expression for the velocity:

$v=\sqrt{\frac{2K}{m}}$

And substitutin into (4), we find:

$r=\frac{\sqrt{2mK}}{qB}$

So, this is the radius of the cyclotron that we must have in order to accelerate the particles at a kinetic energy of K.

Note that for a cyclotron, the acceleration of the particles is achevied in the gap between the dees, where an electric field is applied (in fact, the magnetic field does zero work on the particle, so it does not provide acceleration).

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