DHS 05 Work, Energy & Power (Tutorial Solutions)

2024 H2 Physics Work, Energy, Power Tutorial Solutions 1 | Page _________________________________________________________________________________ Part 1: Work 1 Yes, a body can have energy without having momentum e.g. a stationary book on a table possesses gravitational potential energy. No, a body having momentum will have kinetic energy. 2 Yes, the object is in uniform circular motion. Displacement is always perpendicular to centripetal force. 3 (a) (i) Car A. ma F= Since the same force is applied to both cars, acceleration experienced by car B is smaller (due to its larger mass). 2 2 1 ats = To cover the same distance of 3.0 m, ∆t is larger for car B due to the smaller acceleration. So car A finishes first. (ii) Same work done on both cars and hence same gain in kinetic energy for both cars. (iii) Car B. 22 2 211 22 2 mv pKE mv mm= = = Since the gain in KE is the same for both cars, momentum of car B is larger due to its larger mass. (b) (i) Car A. Since the same force is applied to both cars, acceleration experienced by car B is smaller (due to its larger mass). 2 2 1 ats = With a smaller acceleration, displacement of car B is lesser. (ii) Impulse = Ft p∆= ∆ Hence both cars experienced the same ∆p. (iii) Car A. m pKE 2 2 = Since momentum is the same for both cars, car A has a larger KE due to its smaller mass. 4 (a) The tension in the string does no work because the motion of the pendulum is always perpendicular to the cord and therefore to the tension. (b) The air resistance always does negative work because air resistance is always acting in a direction opposite to the motion. (c) The weight always acts downwards, therefore the work done by the weight is positive on the downswing and negative on the upswing. 5 (a) Since the box is not undergoing acceleration, and the ground is frictionless, the force required to move the box by 5 m is 0 N, Work done = F.s = 0 (b) In order for box to move at constant velocity, the force exerted to keep the box moving must be the same as frictional force. Work done = F.s = 230 (5) = 1150 J

2 | Page o B o 2 B 2 o 2 B2 12 o2 1 B p kA p k v is B at speed the Therefore, v v v v mghmv=mghmv E(E FinalE(E Initial = = ++ += + )) ghv v ghv v g vghv mg mv=mghmv ) E(E Final) E(E Initial 2 o C 2 o 2 C 2 h2 c2 12 o2 1 2 h2 C2 12 o2 1 C p kA p k + = + = + = + ++ += + (c) In order for box to move vertically upwards without an acceleration, the force required is of the same magnitude as the weight of the object. Work done = F.s = 10 (9.81) (4) = 392 J 6 Work done by friction (net external force) on car = change in kinetic energy of car −Fd = 0 − 2 2 1 mv . . (1) −Fdnew = 0 − ( ) 222 1 v m . . (2) (2)/(1): dnew = 4d 7 The bullet loses kinetic energy as it gets stopped in the target. Work done by resistive force on bullet = change in kinetic energy of bullet −F(0.40)= 0 − 2 2 1 mv = 21 (0.300)(500)2− F = 9.4 x