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Question

Figure shows a cord attached to a cart that can slide along a friction less horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height h = 1.20 m, so the cart slides from $$x_1 = 3.00 m$$ to $$x_2 = 1.00 m$$. During the move, the tension in the cord is a constant 25.0 N. What is the change in the kinetic energy of the cart during the move?

Solution
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We solve the problem using the work-kinetic energy theorem, which states that the change in kinetic energy is equal to the work done by the applied force, $$\Delta K = W$$.

In this problem, the work done is $$W = Fd$$, where F is the tension in the cord and d is the length of the cord pulled as the cart slides from $$x_1 $$ to $$x_2$$.

From the figure, we have
$$d = \sqrt{x_1^2 + h^2} - \sqrt{x_2^2 + h^2}$$

$$ = \sqrt{(3.00m)^2 + (1.20 m)^2} - \sqrt{(1.00 m)^2 + (1.20 m)^2} $$

$$= 3.23 m - 1.56 m = 1.67 m$$
which yields $$\Delta K = Fd = (25.0 N)(1.67 m) = 41.7 J.$$

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