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2-25. A block of mass m: I.62 kg slides down a frictionless incline (Figure 2-A). The block is released a height h : 3.91m above the bottom of the loop. (a) What is the force of the inclined track on the block at the bottom (point A)? The track in frictionless and hence the law of conservation of mechanical energy is applicable to the block. According to this law here if the velocity of the block at A is vA then Gain in kinetic energy = loss in potential energy Or Or 1 2 mv A mgh -------------------- (1) 2 vA 2gh 2*9.8*3.91 8.754 m/s Now as the block is moving on a circular track it is having a centripetal acceleration and requires a centripetal force to provide it. If the normal reaction of the track at point A then the resultant force of the weight of the block mg and the normal reaction N at A will provide necessary centripetal force and hence we get mvA2 N mg R N Or the force exerted by the track on the block at point A will be N mg mvA2 v2 m g A R R vA Substituting the value of vA from equation 1 we have mg 2 gh 2h N m g mg 1 R R (b) \t\hat is the force of the track on the block at point B? Now the radius of the track is R and hence the height of point from the ground is R - R cos45o = 1 R 1 2 vB NB Thus the loss in height from the top of the incline will be h = h Or 1 R 1 2 h = h – 0.293 R Hence by applying the law of conservation of mechanical energy between top and point B we have Gain in kinetic energy = loss in potential energy 45o mg Or 1 2 mvB mg h 2 Or vB 2 g h 0.293R m/s Now again the centripetal force towards the center of the track is provided by the resultant of normal reaction at B NB and the component of the weight mg of the block in that direction and hence we have N B mg cos 450 mvB2 R Solving and substituting value of vB from equation 2 we have Or Or mvB2 mg mg N B mg cos 45 2 h 0.293R R R 2 2h 2h N B mg 0.707 0.586 mg 0.121 R R 0 (c) At what speed does the block leave the track? The speed at point B is already calculated in part (b) as vB 2 g h 0.293R m/s (d) How far away from point A does the block land on level ground? After leaving the track at point B the block will move under gravity and makes a projectile motion, projected with speed vB in the direction tangential to the track making angle 450 with the horizontal at a height R sin450 = R from the ground. 2 Applying the formula for the trajectory of a projectile we can calculate the horizontal distance from point of projection for any height (= - R cos450 in this case) as y x tan gx 2 2v02 cos 2 Or R * cos 450 x tan 450 Or R Gives R 2 2 x *1 x *1 gx 2 2vB2 cos 2 450 gx 2 2 * 2 g h 0.293R * 1 2 x2 2 h 0.293R Or 2 h 0.293R R 2 h 0.293R x x 2 Or x 2 2 h 0.293R x 2 R h 0.293R 0 Solving this quadratic we can calculate the value of x. Now x is the horizontal distance of landing point from Point B hence the distance from point A will be x + R sin 45 = x + R 2 (e) Sketch the potential energy U(x) of the block. Indicate the total energy on the sketch. The potential energy of a body at height h with respect to some reference level is given by Potential energy = mgh As acceleration due to gravity g near the surface of earth for h much less then the radius of earth is considered constant and equal to 9.8 m/s2, the potential energy is directly proportional to the height h of the block from the ground (as ground is considered as reference level where the potential is zero). Now the height depends on x according to the shape of the track hence on the track the potential energy graph (verses x) will have the same shape as that of the track and after point B it will be a projectile hence having the shape of a parabola. So the track will be given as U(x) mgh 0 x The initial potential energy is mgh and the total energy is also mgh because at that point the kinetic energy of the block is zero. As the block slides down its kinetic energy increases and the potential energy decreases according to the graph. At the highest point of the parabola also it is having some horizontal velocity and hence the potential energy is less then the total energy. As there is no friction force the total energy remains constant.