Question 9 of 34
1 answer:
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Answer:
7. Net constant force down the ramp
Explanation:
After the car is released and starts moving up the ramp, the only force that is applied on the car is weight because of the gravity, we were told that the friction force is neglected. the force because of the weight is given by:
where θ is the angle of the ramp.
as you can see those values won't change, so the force remains constant down the ramp.
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