A2. A 2.50 kg box, sliding on a rough horizontal surface, has a speed of 1.20 m/s when it makes contact with a spring as shown in Figure A2. The block comes to a momentary stop when the compression of the spring is 5.00 cm. The work done by the friction, from the instant the block makes contact with the spring until is comes to a momentary stop, is -0.500 J. (a) What is the spring constant of the spring? (2 marks) (b) What is the coefficient of kinetic friction between the box and the rough surface? (2 marks)
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First, we need to find the initial kinetic energy of the box before it makes contact with the spring. The formula for kinetic energy is: $KE = \frac{1}{2}mv^2$ where m is the mass of the box (2.50 kg) and v is its speed (1.20 m/s). Show more…
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