Thexand yapplications of Newton's second law provide two equations: F1cos θ– fk– mgsin = maFN– F1sin – mgcos = 0. (a) The trunk is moving up the incline at constant velocity, so a= 0. Using fk= μkFN, we solve for the push-force F1and obtain Fmgkk1=+−sincoscossin.bgThe work done by the push-force GF1as the trunk is pushed through a distance Aup the inclined plane is therefore ()()()()11k23cossincoscoscossin50 kg 9.8 m s6.0 m cos30sin300.20 cos30cos300.20 sin302.2 10 J.kmgWFθθθμ θ+==−°°+°=°−°=×AA(b) The increase in the gravitational potential energy of the trunk is 23sin(50kg)(9.8m/s )(6.0m)sin30
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