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postnew [5]
2 years ago
7

If you start skating down this hill, your potential energy will be converted to kinetic energy. At the bottom of the hill, your

kinetic energy will be equal to your potential energy at the top. What will be your speed at the bottom of the hill?
Physics
1 answer:
baherus [9]2 years ago
6 0
Your potential energy at the top of the hill was (mass) x (gravity) x (height) .

Your kinetic energy at the bottom of the hill is (1/2) x (mass) x (speed)² .

If there was no loss of energy on the way down, then your kinetic energy
at the bottom will be equal to your potential energy at the top.

(1/2) x (mass) x (speed)² = (mass) x (gravity) x (height)

Divide each side by 'mass' :

(1/2) x (speed)² = (gravity) x (height) . . . The answer we get
will be the same for every skater, fat or skinny, heavy or light.
The skater's mass doesn't appear in the equation any more.

Multiply each side by 2 :

(speed)² = 2 x (gravity) x (height)

Take the square root of each side:

<u>Speed at the bottom = square root of(2 x gravity x height of the hill)</u>

We could go one step further, since we know the acceleration of gravity on Earth:

Speed at the bottom = 4.43 x square root of (height of the hill)

This is interesting, because it says that a hill twice as high won't give you
twice the speed at the bottom.  The final speed is only proportional to the
<em>square root </em>of the height, so in order to double your speed, you need to
find a hill that's <em>4 times</em> as high.






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A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height
AleksandrR [38]

a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

E_i=\frac{1}{2}m_1v_1^2

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

v_3 = \frac{m_1 v_2}{m_1 + m_2}=\frac{(60 kg)(5.0 m/s)}{60 kg+15 kg}=4 m/s

c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

E_f = (m_1 +m_2)gh_{max}

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m

4 0
3 years ago
A wagon wheel consists of 8 spokes of uniform diamter, each of mass m, and length L. The outer ring has a mass m rin. What is th
Genrish500 [490]

Answer:

L^2(\dfrac{8m}{3}+m_r)

Explanation:

m = Mass of each rod

L = Length of rod = Radius of ring

m_r = Mass of ring

Moment of inertia of a spoke

\dfrac{mL^2}{3}

For 8 spokes

8\dfrac{mL^2}{3}

Moment of inertia of ring

m_rL^2

Total moment of inertia

8\dfrac{mL^2}{3}+m_rL^2\\\Rightarrow L^2(\dfrac{8m}{3}+m_r)

The moment of inertia of the wheel through an axis through the center and perpendicular to the plane of the ring is L^2(\dfrac{8m}{3}+m_r).

3 0
3 years ago
James and John dive from an overhang into the lake below. James simply drops straight down from the edge. John takes a running s
liraira [26]

Answer:

Both of them reach the lake at the same time.

Explanation:

We have equation of motion s = ut + 0.5at²

Vertical motion of James : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{James}=\sqrt{\frac{2h}{g}}

Vertical motion of John : -

          Initial velocity, u = 0 m/s

         Acceleration, a = g

         Displacement, s = h

    Substituting,

                  s = ut + 0.5 at²

                 h = 0 x t + 0.5 x g x t²

                 t_{John}=\sqrt{\frac{2h}{g}}

So both times are same.

Both of them reach the lake at the same time.

3 0
3 years ago
Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a nega
devlian [24]

Answer: its negative

Explanation: becase it is

5 0
3 years ago
A 2011 Porsche 911 Turbo S goes from 0-27 m/s in 2.5 seconds. What is the car's acceleration?
Natalka [10]

Answer:

-10.8m/s^2

Explanation:

a=change in velocity/change in time

-27 m/s/2.5=10.8m/s^2

or if its not negative

27m/s/2.5=10.8m/s^2

3 0
2 years ago
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