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3 years ago

7

1 kg ball rolls off a 33 m high cliff, and lands 23 m from the base of the cliff. Express the displacement and the gravitational

force in terms of vectors and calculate the work done by the gravitational force. Note that the gravitational force is <0, , 0>, where is a positive number (+9.8 N/kg). (Let the origin be at the base of the cliff, with the direction towards where the ball lands, and the direction taken to be upwards.)

1 answer:

enyata [817]3 years ago

6 0

Answer:

d = <23, 33, 0> m , F_W = <0, -9.8, 0> , W = -323.4 J

Explanation:

We can solve this exercise using projectile launch ratios, for the x-axis the displacement is

x = vox t

Y Axis

y = $v_{oy}$ t - ½ g t²

It's displacement is

d = x i ^ + y j ^ + z k ^

Substituting

d = (23 i ^ + 33 j ^ + 0) m

Using your notation

d = <23, 33, 0> m

The force of gravity is the weight of the body

W = m g

W = 1 9.8 = 9.8 N

In vector notation, in general the upward direction is positive

W = (0 i ^ - 9.8 j ^ + 0K ^) N

W = <0, -9.8, 0>

Work is defined

W = F. dy

W = F dy cos θ

In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º

Cos 180 = -1

W = -F y

W = - 9.8 (33-0)

W = -323.4 J

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Answer:

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1. Energy transferred = 144 Joules.

2. The unit of potential difference, volts can also be described as Joules per Coulombs.

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Explanation:

<u>Part A.</u>

Given the following data;

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Substituting into the equation, we have;

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<u>Part B.</u>

1. <em><u>Given the following data;</u></em>

Charge, Q = 24C

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E = Q * V

Substituting into the equation, we have;

E = 24 * 6

E = 144 Joules.

2. Since we know that, Energy transferred = charge moved * potential difference

$Potential \; difference = \frac {Energy \; transferred}{Charged \; moved}$

The units of energy is Joules while the unit of the quantity of charge moved is Coulomb.

Therefore, the unit of potential difference becomes Joules per Coulomb.

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Q = 27.78C

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Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

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Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

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Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

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$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

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Answer:

143 °

Explanation:

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