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oee [108]
2 years ago
5

The superheroine Xanaxa, who has a mass of 67.5 kg , is pursuing the 70.3 kg archvillain Lexlax. She leaps from the ground to th

e top of a 179 m high building then dives off it and comes to rest at the bottom of a 15.9 m deep excavation where she finds Lexlax and neutralizes him. Does all this bring about a net gain or a net loss of gravitational potential energy?
Physics
1 answer:
Paladinen [302]2 years ago
4 0

Answer:

there is a net loss of potential energy  is Ut = -0.11 10⁵ J

Explanation:

The gravitational potential energy is

        ΔU = mg (y2- y1)

Where y1 is the initial position where the energy is defined as zero, in general in problems on the Earth's surface the reference system is taken at the lowest point, whereby y1 = 0 and consequently U1 = 0, this It is a matter of comfort since what matters are the changes in energy and not its value at one point explained.

Let's apply these concepts to our problem.

Jump from the ground to 179 m, let's calculate the energy change

     ΔU1 = mg (y2-y1)

     ΔU1 = 67.5 9.8 (179 -0)

     ΔU1 = 1.18 10⁵ J           gains energy

Dives up to Y3 = -15.9m below ground, therefore, this height is negative

The total height is y = 179 - (-15.9) = 194.9 m

       ΔU2 = mg (y3-y2)

These are the points where the dive begins and where it arrives

       ΔDU2 = 67.5 9.8 (- 194.9)

       ΔU2 = -1.29 105 J        loses energy

The total change in potential energy is

       Ut = ΔU1 + ΔU2

       Ut = 1.18 10⁵ - 1.29 10⁵

       Ut = -0.11 10⁵ J

Therefore, there is a net loss of potential energy

You can also calculate ΔU = mg (Δy)

Where ΔY is the change in body position, that is -15.9 m

         ΔU = 67.5 9.8 (-15.9)

         ΔU = -1.1 10⁴ J = -0.11 10⁵ J

You can see that it is the same value.

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A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for
stealth61 [152]

Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

(c) x_f=2.6m

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

(c) Using the kinematics equation:

x_f=x_0+v_0t \pm  \frac{at^2}{2}

For the girl, we have x_0=0 and v_0=0. So:

x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

Now, we solve (1) for x_f_g

x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m

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