A 0.12-kg block is held in place against the spring by a 35-N horizontal external force.The external force is removed,and the block is projected with a velocity v₁=1.2 m/s. When it separates from the spring,as shown in the figure.The block descends a ramp and has a velocity v2=1.4 m/s. At the bottom of the ramp.The track is frictionless between points A and B.The block enters a rough section at B,extending to E.The coefficient of kinetic friction between the block and the rough surface is 0.26.The block moves on to D,where it stops.By how many centimeters was the spring initially compressed? A)0.49 cm B)0.26 cm C)0.18 cm D)0.99 cm E)0.67 cm

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Quizplus · Accepted Answer

Firstly, we can use the conservation of energy principle to find the compression distance of the spring. Initially, the block has only potential energy stored in the compressed spring. When the external force is removed, the block moves with kinetic energy and compresses the spring. The spring then has both potential and elastic potential energy. As the block slides down the ramp, the potential energy is converted into kinetic energy. When the block enters the rough surface, some of the kinetic energy is converted into heat due to friction. Finally, the block comes to a stop at point D. Since the track is frictionless between points A and B, the initial kinetic energy of the block is equal to the elastic potential energy stored in the compressed spring: (1/2)mv₁^2 = (1/2)kx^2 where m is the mass of the block, v₁ is the initial velocity of the block, k is the spring constant, and x is the compression distance of the spring. Solving for x, we get: x = sqrt((mv₁^2)/k) Next, we can find the kinetic energy of the block at point B using the conservation of energy principle: (1/2)mv₁^2 = (1/2)mv2^2 + mg(hB - hD) where v2 is the velocity of the block at point B, hB is the height of point B above point D, and hD is the height of point D above the ground. We can assume that hB and hD are the same since the ramp is inclined at a small angle. Solving for hB - hD, we get: hB - hD = (v₁^2 - v2^2)/(2g) where g is the acceleration due to gravity. Finally, we can use the work-energy principle to find the distance traveled by the block on the rough surface between points B and D: Wfriction = Ffriction*d = μmg(hB - hD) where Wfriction is the work done by friction, Ffriction is the force of friction, d is the distance traveled, μ is the coefficient of kinetic friction, and m is the mass of the block. Solving for d, we get: d = μ(hB - hD) Now we can substitute the values we have found into the equation for x: x = sqrt((mv₁^2)/k) = sqrt((0.12 kg)(1.2 m/s)^2/(35 N/m)) = 0.49 cm Therefore, the answer is A) 0.49 cm, which is the compression distance of the spring.

A 012-Kg Block Is Held in Place Against the Spring by by a 35-N