# Acar on a roller coaster starts at zero speed at an elevation above the ground of 26 m. it coasts down

Acar on a roller coaster starts at zero speed at an elevation above the ground of 26 m. it coasts down a slope, and then climbs a hill. the top of the hill is at an elevation of 16 m. what is the speed of the car at the top of the hill? neglect any frictional effects. [hint: use conservation of mechanical energy]

## This Post Has 5 Comments

1. microwave13016 says:

14 m/s

Explanation:

The velocity u at the ground from an elevation of 26 m is given by

$u=\sqrt{2gh} =\sqrt{2\times9.8\times26} \text{ m/s}$

Now, the speed of the car v at the top of the hill , which is at an elevation of 16 m can be calculated by

$v^2 =u^2-2gh= 2\times9.8\times26-2\times9.8\times16\\=2\times9.8\times10\\=196\\=14 \text{ m/s}$

2. Dragonflare says:

The speed of the car at the top of the hill is 14m/s

Explanation:

given that

Initial velocity u of the car=0 m/s

The distance can be determined by finding out the difference between the elevation of the first slope and second slope.

elevation of the first slope=26 m

elevation of second slope=16m

distance s=26-16=10 m

acceleration due to gravity g=9.8 m/s2

speed of the car at the top of the hill can be determined by using the equation

$v^2=u^2+2as\\v^2=0^2+2\times 9.8\times 10\\=196\\v=\sqrt{196} =14m/s$

speed of the car at the top of the hill is 14m/s

3. desstinee863 says:

option (c)

Explanation:

h1 = 26 m

h2 = 16 m

Let the speed of the car is v.

Use the law of energy conservation

Potential energy at initial point = Potential energy at final point + Kinetic energy at final point

m x g x h1 = m x g x h2 + 1/2 x m x v^2

9.8 x 26 - 9.8 x 16 = 0.5 x v^2

196 = v^2

v = 14 m/s

4. angelaOR says:

v = 14 m / s

Explanation:

For this exercise Let's use the concepts of mechanical energy. Let's write in two points, the highest one where energy is potential and the lowest one where energy is potential and kinetic part pate

initial. Highest

Em₀ = U = mg y₁

final. top  on the hill

$E_{mf}$ = K + U = ½ m v² + mg y₂

As there is no rubbing

Em₀ = $E_{mf}$

mg y₁ = ½ m v² + mg y₂

v = √ 2g (y₁ -y₂)

Let's calculate

v = √ 2 9.8 (26 - 16)

v = 14 m / s

5. clevelandjaniya says:

The speed of the car at the top of the hill is C) 14 m/s

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Further explanation

Let's recall the formula of Kinetic Energy as follows:

$\large {\boxed {E_k = \frac{1}{2}mv^2 }$

Ek = Kinetic Energy ( Newton )

m = Object's Mass ( kg )

v = Speed of Object ( m/s )

Let us now tackle the problem !

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Given:

initial height = h₁ = 26 m

final height = h₂ = 16 m

initial speed = v₁ = 0 m/s

Asked:

final speed = v₂ = ?

Solution:

We will use Conservation of Energy formula to solve this problem as follows:

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$mgh_1 + \frac{1}{2}m(v_1)^2 = mgh_2 + \frac{1}{2}m(v_2)^2$

$mgh_1 + \frac{1}{2}m(0)^2 = mgh_2 + \frac{1}{2}m(v_2)^2$

$mgh_1 = mgh_2 + \frac{1}{2}m(v_2)^2$

$gh_1 = gh_2 + \frac{1}{2}(v_2)^2$

$2g(h_1 - h_2) = (v_2)^2$

$v_2 = \sqrt {2g(h_1 - h_2)}$

$v_2 = \sqrt { 2(9.8)(26 - 16) }$

$v_2 = \sqrt { 196 }$

$v_2 = 14 \texttt{ m/s}$

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Learn moreImpacts of Gravity : Effect of Earth’s Gravity on Objects : The Acceleration Due To Gravity : Newton's Law of Motion: Example of Newton's Law:

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

$Acar on a roller coaster starts at zero speed at an elevation above the ground of 26 m. it coasts do$