Physic Labs

Problem 2

A block of mass m=2.0 kgm=2.0\,\mathrm{kg} starts from rest and slides s=5.0 ms=5.0\,\mathrm m down a 45∘45^\circ incline. The kinetic-friction coefficient is μ=0.20\mu=0.20; take g=10 m s−2g=10\,\mathrm{m\,s^{-2}}. The plane is fixed and the block remains in contact with it throughout the slide. The coefficient μ\mu is constant along the path, and kinetic friction opposes the motion. The block moves in a straight line down the plane with constant acceleration from its starting point to the stated endpoint. (a) Find the normal force and the acceleration along the plane. (b) Determine the speed at the end of the descent. (c) Calculate the time for the slide, assuming constant acceleration.
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ma=mgsin⁡45∘−f,a=g(sin⁡45∘−μcos⁡45∘)ma=mg\sin45^\circ-f,\quad a=g(\sin45^\circ-\mu\cos45^\circ)
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Along the plane, gravity pulls downhill while friction opposes the sliding. Newton’s law therefore gives the net acceleration as g(sin⁡θ−μcos⁡θ)g(\sin\theta-\mu\cos\theta).

problems.proof.pitfall. Friction points up the plane as the block slides down; compute f=μNf=\mu N, not μmg\mu mg.