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2.0-kilogram block sliding down a ramp starts from rest from a height of 3.0 meters above the ground reaches the ground with a maximum speed of 2.6 m/s. What is the total work done by friction on the block as it slides down the entire length of the ramp?

User Prahack
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Answer:

Explanation:To solve this problem, we can use the work-energy theorem, which states that the net work done on an object is equal to its change in kinetic energy.At the top of the ramp, the block has potential energy, which is given by:PE = mghwhere m is the mass of the block, g is the acceleration due to gravity, and h is the height of the ramp. Substituting the given values, we have:PE = (2.0 kg)(9.81 m/s^2)(3.0 m) = 58.86 JAt the bottom of the ramp, the block has kinetic energy, which is given by:KE = (1/2)mv^2where v is the speed of the block at the bottom of the ramp. Substituting the given values, we have:KE = (1/2)(2.0 kg)(2.6 m/s)^2 = 6.76 JThe net work done on the block is equal to the change in its kinetic energy, which is:W_net = KE - PEW_net = 6.76 J - 58.86 JW_net = -52.1 JSince the block is slowing down due to friction, the work done by friction is negative. Therefore, the total work done by friction on the block as it slides down the entire length of the ramp is:W_friction = -(-52.1 J)W_friction = 52.1 JTherefore, the total work done by friction on the block is 52.1 J.

User Average Kid
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