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kotykmax [81]
3 years ago
14

A block with mass of 10 kg is on a frictionless surface. One hand on the left side of the block is pushing it to the right. A se

cond hand on the right side of the block is pushing it to the left. The block starts from rest. Then Hand 1 pushes with a force of 8 N and the block moves to the right a distance of 4 m, where it has a final velocity of 2 m/s. Between its initial and final positions, how much work did Hand 2 do?
Physics
1 answer:
igomit [66]3 years ago
6 0

Answer:

W_2=-12J

Explanation:

The work of force 2 will be given by the vectorial equation W_2=F_2.d. We know the value of F_1 and have information about its movement, which relates to the net force F=F_1+F_2.

About this movement we can obtain the acceleration using the equation v_f^2=v_i^2+2ad. Since it departs from rest we have a=\frac{v_f^2}{2d}.

And then using Newton's 2dn Law we can obtain the net force F=ma, thus we will have F_2=F-F_1=ma-F1=\frac{mv_f^2}{2d}-F_1

And we had the work done by force 2 as:

W_2=F_2.d=\frac{mv_f^2}{2}-F_1d

(The sign will be given algebraically since we take positive the direction to the right.)

With our values:

W_2=\frac{(10kg)(2m/s)^2}{2}-(8N)(4m)=-12J

<em>Another (shorter but maybe less intuitive way for someone who is learning) way of doing this would have been to say that the work done by both forces would be equal to the variation of kinetic energy:</em>

<em>W_1+W_2=K_f-K_i=K_f=\frac{mv_f^2}{2}</em>

<em>Which leads us to the previous equation straightforwardly.</em>

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Answer:

 Δt = 5.85 s

Explanation:

For this exercise let's use Faraday's Law

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The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.

suppose a linear change of the magnetic field

            emf = - A \frac{B_f - B_o}{ \Delta t}

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4 0
3 years ago
A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s
klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

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The energy stored in the inductor is given by

U = \frac{1}{2} Li^{2}

The rate at which the energy is being stored in the inductor is given by

\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1

The current through the RL circuit is given by

i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })

Where τ is the the time constant and is given by

\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16

i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}

Therefore, eq. 1 becomes

\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})

At t = 0.13 seconds

\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts

(b) thermal energy is appearing in the resistance

The thermal energy is given by

P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts

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The energy delivered by battery is

P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts

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Answer:

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Answer

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8 0
3 years ago
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