Posted by Sphelele on .
Abox of mass 60kg starts from the rest at height h and slides down a rough slope of length 10m, which makes an angle of 25 degree with the horizontal. It undergoes a constant accelequation of magnitude 2m/ while sliding down the slope. Calculate the work done on the box by the frictional force, using the work-energy theorem
Wb = m*g = 60kg * 9.8N/kg = 588 N. = Wt.
Fp = 588*sin25 = 248.5 N. = Force
parallel to the slope.
Fp-Fk = m*a
248.5-Fk = 60 * 2 = 120
Fk = 248.5 - 120 = 128.5 N. = Force of
Work = Fk * L = 128.5 * 10 = 1285 Joules