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anzhelika [568]

3 years ago

12

A roller coaster starts from rest at its highest point and then descends on its (frictionless) track. its speed is 40 m/s when i

t reaches ground level. part a what was its speed when its height was half that of its starting point? what was its speed when its height was half that of its starting point? 33 m/s 14 m/s 20 m/s 28 m/s

1 answer:

Alenkinab [10]3 years ago

5 0

Given that,

Initial velocity , Vi = 0

Final velocity , Vf = 40 m/s

Acceleration due to gravity , a = 9.81 m/s²

Distance can be calculated as,

2as = Vf² - Vi²

2 * 9.81 *s = 40² - 0²

s = 81.55 m

For half height, that is, s = 40.77m

Vf= ??

2as = Vf² - Vi²

2 * 9.81 * 40.77 = Vf² - 0²

Vf² = 800

Vf = 28.28 m/s

Therefore, speed of roller coaster when height is half of its starting point will be 28 m/s.

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a. The spring is compressed by 0.174 m.

b. The vertical distance the dart travels from its position when the spring is compressed to its highest position is 9.6 m.

c. The horizontal velocity of the dart at that time is 13.74 m/s.

d. The horizontal distance from the equilibrium position at which the dart hits the ground is 19.236 m.

<h3>What is potential energy?</h3>

The energy by virtue of its position is called the potential energy.

a. Given is the potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J.

PE of spring = 1/2 kx²

Put the values, we get

P.E = 0.940 = 1/2 x 62 x²

x = 0.174 m

Thus, the spring is compressed by 0.174 m.

b. Given is a 0.010 kg dart is fired straight up.

The vertical height is find out by

0.940 J = (0.010 kg) (9.8 m/s²) h

h = 9.6 m

Thus, the vertical distance the dart travels from its position is 9.6 m

c. From the conservation of energy principle, total mechanical energy is conserved.

1/2 mv² =mgh

v = √2gh

Plug the values, we get

v = √2 x 9.8x 9.6

v = 13.74 m/s

Thus, the horizontal velocity is 13.74 m/s.

d. Time that dart spends in air, t = √2h/g

t = √(2x9.6)/9.81

t = 1.4 s

The horizontal distance from the equilibrium position at which the dart hits the ground.

Horizontal distance = (Velocity on x direction) x time

Horizontal distance = 13.74 m/s x 1.4s

Horizontal distance = 19.236 m

Thus, the horizontal distance is 19.236 m.

Learn more about potential energy.

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a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T

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Explanation:

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$T=\sqrt{\frac{m}{F}}$

Integrating to get $\dot{x}$ with initial conditions $\dot{x}(0)=0$ :

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}$

Integrating to get x with initial conditions x(0) = 0:

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When t=T:

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Answer: $0.084 m/s^2$

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