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3 years ago

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You and a friend each carry a 15 kg suitcase up two flights of stairs, walking at a constant speed. Take each suitcase to be the

system. Suppose you carry your suitcase up the stairs in 30 s while your friend takes the 60s.
Which of the following is true?
a. You did more work, but both of you expended the same power.
b. You did more work and expended more power.
c. Both of you did the same work, but you expended more power.
d. Both of you did the same work, but you expended less power.

1 answer:

AlekseyPX3 years ago

7 0

Answer:

Both of you did the same work but you expended more power.

Explanation:

<em>Work done</em> by an object is calculated by force applied multiplied by the distance.

W=F*d

From the figure given below let us calculate force applied bith you and yopur friend.

Let us take the stairs in positive x direction,

Work done by you W₁ ,

The force applied Fₓ = F - mgsinθ =maₓ

here aₓ = 0, because both of you move with constant speed

F - mgsinθ = 0

F= mgsinθ

The work done by you on the suitcase is

W = F L cos0° , where L is he length of the staircase.

W = FL = mgsinθL , by substituting value of F

Work done by you is W₁ = mgLsinθ

Similarly work done by your friend is W₂ = mgLsinθ.

Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .

Therefore <em>work done by both of you is same</em> . Both of you did equal work.

The power , is defined as amount of energy converted or transfered per second or rate at which work is done .

P = $\frac{W}{t}$ = $\frac{FL}{t}$

Power spend by you P₁ = mgLsinθ/t

P₁ = 15*9.8*Lsinθ/30

P₁ = 4.9L sinθ eqn 1

Power spend by your friend is P₂ = mgLsinθ/t

P₂ =15*9.8*Lsinθ/60

P₂ = 2.45Lsinθ eqn 2

Dividing eqn 1 and eqn 2

P₁ = 2P₂

You have spend more power than your friend .

Hence Both of you did equal work but you spend more power.

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Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

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Answer:

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The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

a)

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

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$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

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