JPJC 2026 Energy and Fields Tutorial Solutions

2026/JPJC/PHYSICS/9478 1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS ENERGY AND FIELDS TUTORIAL SOLUTIONS Qn. Solution Self-Check Questions S1 The work done by a force on an object is defined as the product of the force and the displacement of the object in the direction of the force. S2  Kinetic energy is a scalar quantity that represents the energy associated with the body due to its motion.  Consider an object of mass m on a horizontal frictionless surface. A constant force F acts on the object, causing it to accelerate from an initial velocity of u to a final velocity v, covering a displacement of s.  Work done by force F on the object is transferred to the object as its kinetic energy (law of conservation of energy). An equation for kinetic energy would need to link work done to the velocity of the object. This can be done by applying the kinematics equations.  Work done by force F on the object, FsW  - - - (1)  Since the surface is frictionless, F is the net force acting on the mass. Hence F in equation (1) can be replaced by F ma . ( ) - - - (2) W Fs W ma s    Since the net force is constant, acceleration a is constant, and the kinematics equation of motion applies.  2 2 2v u as )(2 1 22 uvsa  - - - (3)  Substitute (3) into (2), work done by force F is: v m m F F s u Fig. 10: Object accelerating under constant force

2026/JPJC/PHYSICS/9478 2 Qn. Solution 22 22 22 2 1 2 1 )(2 1 )(2 1m )( mumvW uvmW suvsW smaW          Work-Energy Theorem states that: The net work done by external forces acting on an object (or a system) is equal to the object’s (or system’s) change in kinetic energy. k final k initial k work done by net force, ( ) ( ) W E E W E      If the object accelerates from rest to a final velocity of v, all the work done is transferred into the object’s kinetic energy Ek : S3  Near the Earth’s surface, the gravitational field is uniform and the acceleration of free fall g is taken to be constant.  Consider an object of mass m lifted by a constant force F, vertically upwards at constant speed from ground level to a height h .  At constant speed, the applied force must balance the weight of object, F mg  Since the object’s velocity is constant, its kinetic energy is also constant. 2 k 1 2E mv mg F mg F Fig. 11: object rising at constant velocity

2026/JPJC/PHYSICS/9478 3 Qn. Solution  Hence work done by the applied force F on the object is transferred to the object to increase its gravitational potential energy (law of conservation of energy).  Work done by F is W Fs W mg h    pE mg h   S4 Efficiency of a practical device is a measure of the useful energy output to the total energy input. S5 The gravitational field strength at a point in space is the gravitational force experienced per unit mass on a mass at that point. G 2 F GMg m r   S6 The electric field strength at a point is the electric force exerted per unit charge on a positive charge placed at that point. E 2 04 F QE q r   S7 Power is defined as the rate at which work is done.  Consider a constant force F being applied on an object such that F cancels other forces on the object, resulting in a net force of zero on the object. Under the influence of this force F, the object moves at a constant velocity of v and cover a displacement s over a time interval t. Refer to Fig. 1. Fig. 1: Work done by a constant force moving object at constant velocity  Work done by the force F, W = Fs  Power delivered by the force F,  useful energy output= ×100%total energy input

2026/JPJC/PHYSICS/9478 4 Qn. Solution WP t FsP t sP F t P Fv         Qn. Solution Self-Practice Questions P1(a)    cos 200 10 cos30 1732 1700 J W Fs      P1(b) No work is done as the force exerted is perpendicular to the direction of motion of the satellite. P1(c)    cos 800 0.5 cos180 400 J W Fs     P2 Using the principle of conservation of energy, Loss of gravitational PE = Gain of KE + thermal energy thermal energy        2 2 1 2 120 9.81 4.0 20 2.52 722 J mgh mv     4.0 m u = 0 m s−1 v = 2.5 m s−1

2026/JPJC/PHYSICS/9478 5 Qn. Solution P3(a) P3(b) P4    total WAverage power, totaltime 8000 20 4.0 40000 W 40 kW P Fs t      P5 At constant speed, driving force F = frictional force f 2 3P Fv fv kv v kv     (where k is a constant) 3 2 2 1 1 3 2 40 (26) 208 kW20 P v P v P          

2026/JPJC/PHYSICS/9478 6 Qn. Solution Discussion Questions 1 Work done by the spring during the reduction in length from l1 to l2 = area under graph during reduction in length from l1 to l2 = area MNQP 2(a) The kinetic energy store of the bungee jumper at the start and the end of the jump is zero. Using the principle of conservation of energy, Decrease in GPE = gain in EPE   2 1 160.0 9.81 31.0 31.0 12.02 101.1 101 N m k k          2(b) When the jumper reaches the lowest point, tension in the cord, 101.1 19.0 1920.9 1921 N T kx     Taking upwards as positive,    2 60.0 1920.9 60.0 9.81 22.21 22.2 m s ma T mg a a        3(a) Length of ramp 2 210.0 24.0 26.0 m   By Conservation of Energy, Work done on crate + Loss in GPE = Gain in KE         2 1 140.0 26.0 15.0 9.81 10.0 15.0 2 18.299 18.3 m s v v      

2026/JPJC/PHYSICS/9478 7 Qn. Solution 3(b) By Conservation of Energy, Gain in PEelastic + Work done against friction = Loss in KE      221 1200 100 15.0 18.32 2 4.5359 4.54 m x x x     4(a)(i) Loss in GPE 0.400 9.81 0.200 0.7848 0.785 J mgh      4(a)(ii) At equilibrium, 1 0.200 0.400 9.81 19.62 N m kx mg k k      Gain in EPE    2 2 1 2 1 19.62 0.2002 0.3924 0.392 J kx    4(b) gravitational potential energy / J elastic potential energy / J kinetic energy / J total energy / J 0.200m below point X 0 1.57 0 1.57 point X 0.785 0.392 0.392 1.57 0.200m above point X 1.57 0 0 1.57 5(a) GPE lost by bird = KE gained by bird m(9.81)h = 1 2 m(9.0)2 h = 4.13 m

2026/JPJC/PHYSICS/9478 8 Qn. Solution 5(b)(i) KE at P    23 1 1 120 10 13.02 10.14 10.1 m s       5(b)(ii) Total gain in GPE     3 1 120 10 9.81 7.5 8.829 8.83 m s       5(b)(iii) GPE gained by bird = KE lost by bird (Taking GPE = 0 at P) mgh – 0 = KE at P – KE at nest 8.83 = 10.1 – KE at nest KE at nest = 1.27 J 5(b)(iv) KE of bird as it reaches its nest   3 2 1 1 120 10 1.272 4.60 m s v v      6(a) Free Body Diagram: N Motion on flat ground: Since speed is constant, acceleration = 0 m s−2 Hence resultant force on car = 0 N If F is engine force, flat flat 200 0 200 N F F    200N F(flat) F(incline) 200N 8° N 9810 N 9810 N

2026/JPJC/PHYSICS/9478 9 Qn. Solution flat flat 3 200 20 4000 4.0 10 W P F v       6(b) Motion up an incline:       incline incline 200 1000 9.81 sin8 0 200 1000 9.81 sin8 1565 N F F         incline incline 4 1565 20 31300 3.1 10 W P F v        7(a) Work done against gravity 3 50.0 9.81 8.00 3924 3.92 10 J mg h        7(b) Average power output = work done / time 3924 20 196.2 196 W    7(c)   power outputefficiency power input 196.2power input 0.8 2