1. kulwant Work-Energy Principle Conservation of Mechanical Energy Work done = Change in K.E. + Change in PE A rough slope is inclined at tan -1 ( ¾ ) to the horizontal. A box of mass 25 kg slides from rest a distance of 20 m down a line of greatest slope, at which point it has developed a speed of 14 m/s. Find the work done against friction and the coefficient of friction. Gain in K.E. = 0.5  25  14 2 = 2450 Loss in P.E. = 25  9.8  20  3 / 5 = 2940 Loss in energy = 490 = WD against friction

2. kulwant Work-Energy Principle Loss in energy = 490J = WD against friction WD = F  d Friction = 490  20 = 24.5 N  = 24. 5  25  9.8  ( 4 / 5 ) = 0.125

3. kulwant Work-Energy Principle A bullet of mass 20 grams is fired horizontal into a vertical target of 15 cm thickness with a speed 250 m/s. If the target offers a resistance of 3600N, with what speed will the bullet emerge on the far side? K.E. = 0.5  0.020  250 2 = 625 J WD against resistance = 3600  0.15 = 540 J Energy level on far side = 85 v = 92.2 m/s