A block is projected up a frictionless inclined plane with initial speed v1 = 3.45 m/s. The angle of incline is 𝜃 = 31.7°.(a) How far up the plane does it go? m(b) How long does it take to get there? s(c) What is its speed when it gets back to the bottom? m/sSection 6.6 Applying Newton's Laws
Question
A block is projected up a frictionless inclined plane with initial speed v1 = 3.45 m/s. The angle of incline is 𝜃 = 31.7°.(a) How far up the plane does it go? m(b) How long does it take to get there? s(c) What is its speed when it gets back to the bottom? m/sSection 6.6 Applying Newton's Laws
Solution
To solve this problem, we need to use the principles of physics, specifically kinematics and Newton's laws.
(a) How far up the plane does it go?
We can use the equation of motion: v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.
Since the block is projected up the plane, it will eventually come to a stop, so v = 0. The acceleration is the component of gravity acting down the slope, which is -g*sin(𝜃), where g is the acceleration due to gravity (9.8 m/s²).
So, we can rearrange the equation to solve for s: s = (v² - u²) / 2a
Substituting the given values: s = (0 - (3.45 m/s)²) / (2 * -9.8 m/s² * sin(31.7°)) = 0.61 m
(b) How long does it take to get there?
We can use the equation of motion: v = u + at, where t is the time.
Rearranging to solve for t: t = (v - u) / a
Substituting the given values: t = (0 - 3.45 m/s) / (-9.8 m/s² * sin(31.7°)) = 0.71 s
(c) What is its speed when it gets back to the bottom?
The speed of the block when it gets back to the bottom will be the same as the initial speed, due to the conservation of energy. So, the speed is 3.45 m/s.
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