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Otrada [13]
3 years ago
13

What’s the kinetic energy of the roller coaster at the top and bottom of the hill? Use . A kiddie roller coaster car has a mass

100 kilograms. At the top of a hill, it’s moving at a speed of 3 meters/second. After reaching the bottom of the hill, its speed doubles. The car’s kinetic energy at the bottom is its kinetic energy at the top. The car has joules of kinetic energy at the bottom of the hill.
Physics
2 answers:
serious [3.7K]3 years ago
0 0
K.E1=1/2×100×3²
=50×9
=450J
K.E2=1/2×100×36
=50×36
=1800J
Ymorist [56]3 years ago
0 0

double, 1800

if the speed doubles ,then so does the kinetic energy and 3 would turn into 6

Now we use KE=1/2mv^2

1/2 x 100= 50 x 6^2 = 1800


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A thin rod of length 0.75 m and mass 0.42 kg is suspended
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Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

              d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

              I = ⅓ m L²

we substitute

            w = \sqrt{\frac{mgL}{2}  \ \frac{1}{\frac{1}{3} mL^2} }

            w = \sqrt{\frac{3}{2}  \ \frac{g}{L} }

            w = \sqrt{ \frac{3}{2} \ \frac{9.8}{0.75}  }

            w = 4.427 rad / s

an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

             w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

           w = 4.0 rad / s

the kinetic energy is

           K = ½ I w²

           K = ½ (⅓ m L²) w²

           K = 1/6 m L² w²

           K = 1/6 0.42 0.75² 4.0²

           K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

             Em₀ = K = ½ I w²

final point. Highest point

             Em_f = U = m g h

energy is conserved

             Em₀ = Em_f

             ½ I w² = m g h

             ½ (⅓ m L²) w² = m g h

             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

             h = 0.153 m

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