A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force along an x axis is applied to the block. The force is given by F→ (x) = (2.5 − x 2 ) ˆ i N, where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 2.0 m? (b) What is the maximumkinetic energy of the block between x = 0 and x = 2.0 m?
Question
A 1.5 kg block is initially at rest on a horizontal frictionless surface when a horizontal force along an x axis is applied to the block. The force is given by F→ (x) = (2.5 − x 2 ) ˆ i N, where x is in meters and the initial position of the block is x = 0. (a) What is the kinetic energy of the block as it passes through x = 2.0 m? (b) What is the maximumkinetic energy of the block between x = 0 and x = 2.0 m?
Solution
To solve this problem, we need to use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
(a) The work done on the block as it moves from x = 0 to x = 2.0 m is given by the integral of the force with respect to displacement. The force is a function of position, F(x) = 2.5 - x^2. So, the work done is:
W = ∫ F dx from 0 to 2.0 = ∫ (2.5 - x^2) dx from 0 to 2.0 = [2.5x - (x^3)/3] from 0 to 2.0 = 2.5*2 - (2^3)/3 - 0 = 5 - 8/3 = 7/3 J.
Since the block starts from rest, its initial kinetic energy is zero. Therefore, by the work-energy theorem, its kinetic energy at x = 2.0 m is equal to the work done on it, which is 7/3 J.
(b) The force is maximum at x = 0 and decreases as x increases. Therefore, the work done on the block, and hence its kinetic energy, is maximum at x = 0. As the block moves from x = 0 to x = 2.0 m, the work done on it decreases because of the decreasing force. Therefore, the maximum kinetic energy of the block between x = 0 and x = 2.0 m is also 7/3 J.
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