IS IT CORRECT ?? PLSS
1 answer:
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Answer:
Explanation:
Let y₀ be the initial position of the cylinder when the spring is attached and y its position when it is momentarily at rest.From work-kinetic energy principles, The work done by the spring force + work done by friction + work done by gravity = kinetic energy change of the cylinder
work done by the spring force = ¹/₂k(y₀² - y²)
work done by friction = - f(y - y₀)
work done by gravity = mg(y - y₀)
kinetic energy change of the cylinder = ¹/₂m(v₁² - v₀²)
So ¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(v₁² - v₀²)
Since the cylinder starts at rest, v₀ = 0. Also, when it is momentarily at rest, v₁ = 0
¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = ¹/₂m(0² - 0²)
¹/₂k(y₀² - y²) - f(y - y₀) + mg(y - y₀) = 0
¹/₂ky₀² + fy₀ - mgy₀ -¹/₂ky² - fy + mgy = 0
¹/₂ky₀² + fy₀ - mgy₀ = ¹/₂ky² + fy - mgy
Let y₀ = 0, then the left hand side of the equation equals zero. So,
0 = ¹/₂ky² + fy - mgy
¹/₂ky² + fy - mgy = 0
Using the quadratic formula
Answer:
3.0 seconds
Explanation:
The time of flight of a projectile (the time it takes to reach the ground) does not depend on the horizontal motion, but only on its vertical motion.
In fact, the time of flight is determined by the suvat equation:
where
s is the vertical displacement
u is the initial vertical velocity (0, in case of these two projectiles)
g = 9.8 m/s^2 is the acceleration of gravity (assuming downward as positive direction)
t is the time of flight
Re-arranging the equation, we get
We see that this time depends only on s (the heigth of the cliff) and g: therefore, since the two projectiles are launched from the same height, they take the same time to reach the ground, 3.0 seconds.