1 Dynamics Worksheet Answers (a) Answers: A vector quantity has direction while a scalar quantity does not have direction. Answers: (D) Velocity, weight and friction are vector quantities. Note: weight and friction are forces, which are vector quantities. Answers: (A) Mass, energy, power and time are scalar quantities which do not have direction. (b)
2 Answers: (C) When a body s in equilibrium, the resultant force is zero. The tip-to-tail force diagram starts from Q and return to starting point of Q. Therefore, the resultant force is zero.
5 (c) Answers: (C) Since the speed is constant, hence acceleration is zero. By Newton s First Law, the forces are balanced. Hence resultant force is zero. From the force diagram above, resultant force, Fr = 1500 N F. Since Fr = 0; 1500 N F = 0, hence, F = 1500 N. Answers: Given
6 Since the speed is constant, the acceleration is zero. By Newton s First Law, the forces are balanced. Hence the resultant force is zero. Thus there is a friction force present, acting in the opposite direction of the 10 N force. Since the forces are balanced, the friction between the block and the contact surface equals the applied force (10 N) to give zero resultant force; illustration as shown in the diagram below. New situation, the applied force is 18 N now. The friction force Fr remains as 10 N. From the force diagram above, the resultant force = 18 N Fr = 18 N - 10 N = 8 N Use F = ma to find the acceleration of the block, given resultant force, F = 8 N, m = 4 kg a = F / m = 8 / 4 = 2 m/s 2
7 Answers: (C) Explanation: The resultant force = 1200 N 800 N = 400 N (upwards)
8 By Newton s Second Law, F = ma, the acceleration is in the same direction of the resultant force, hence the acceleration is upwards. While the parachutist is moving downwards, the upward resultant force causes it to decelerate (slow down). Since air resistance force depends on the speed, the air resistance decreases as the speed of the parachutist decreases. The air resistance will decrease until it is the same as the weight of the parachutist. AT this point of time, there forces are balanced. Resultant force is zero. Acceleration is zero. The parachutist descents at a constant speed. Answers: (D)
9 By Newton s Third Law (action = reaction), the tension force acting on the man equal the tension force acting on the girl. Thus Fm = Fg. Assuming the fat man has a larger mass than the thin girl. By Newton s Second Law, F=ma, for the same amount of force action on both fat man and thin girl, the person with a smaller mass will have a larger acceleration. This the thin girl accelerates more. (d) Answer: (A) As a result of the opposing force, the object will experience a balanced force. By Newton s First Law, the object will continue with its uniform motion in straight line. Since the object is already accelerating (with a certain speed), it will continue to move at a constant speed it has attained before the opposite force acts on it.
10 (e) Answer: (C) The smallest magnitude of the resultant force is 6 4 = 2 N The largest magnitude of the resultant force is = 10 N The two forces can be in any direct ions to form a resultant force of 5 N. See example below: Answers: (A) Draw a free body diagram to indicate all the forces involved.
11 From the free body diagram, the resultant force = 30 - F r ---- (1) Given mass, m = 5 kg, acceleration, a = 4 m/s 2 By Newton s Second Law, resultant force, F = ma = 5 x 4 F = 20 N ---- (2) From (1) and (2), 30 F r = 20 N Fr = 10 N Answers: (A) The smallest magnitude of the resultant force = = 20 N.
12 Answers: (C) Since the object moves at constant velocity, by Newton s First Law, the forces are balances. Hence the friction force is present and has the same value as the applied force, 80N By Newton s First Law, F r = 80 N Draw a free body diagram to indicate all forces involved. From the free body diagram, resultant force = F A 80 N (1) By Newton s Second Law, resultant force, F = ma = 20 kg x 0.4 m/s 2
13 = 8 N (2) From (1) & (2) F A 80 N = 8 N F A = = 88 N (g) Answer: (D) By Newton s Second Law, F = ma, the acceleration of the car will increase in the first part of the journey.
14 The gradient of the velocity-time graph equals to the acceleration. Since the acceleration is increasing, the gradient of the velocity-time graph will also be increasing. At the second part of the journey, F is constant; hence the acceleration is also constant. Therefore the gradient of the velocity-time graph is also constant. Answer: (B) Answer: (D) Method 1 By Newton s Second Law, F = ma, the resultant force required to accelerate the mass at 5 m/s 2 F = 10 x 5 = 50 N. Since friction force (30 N) is opposing the applied force, the applied force must be 30 N greater than the resultant force. 30 N greater than 50 N is 80 N. Method 2
15 See the free body diagram below for illustration. From the free body diagram, the resultant force = F A 30 N (1) By Newton s Second Law, F = ma, the resultant force required, F = 10 x 5, F = 50 N (2) From (1) & (2), F A 30 N = 50 N F A = = 80 N. Answer: (C) Method 1 By Newton s Second Law, F = ma, when the mass is reduced by half, the resultant required is also reduced by half. Therefore F = 15 N. When the acceleration increases by 3 times, the resultant force must also increase by 3 times. Therefore, the final resultant force is 3 x 15 N = 45 N
16 Method 2 By Newton s Second Law, F A = M x A F B = ½ M x 3A Take the ratio of F B / F A, F B / F A = MA / 3 x ½ MA F B = 3/2 F A Since F A = 30 N, F B = 3/2 (30) = 45 N
17 Answer: (B) Determine the magnitude of the friction force F R. Use a free body diagram to help visualize the forces involved. By Newton s Second Law, F = ma, the resultant force required to accelerate the 2 kg mass is F = 2 x 3 = 6 N (1) From the free body diagram, The resultant force is 10 F R (2) From (1) & (2) 10 F R = 6 N F R = 4 N
18 Now a force of 5 N is applied in the opposite direction. Since the resultant force has changed, the acceleration of the mass will change. From the free body diagram, resultant force = = 1 N By Newton s Second Law, F = ma, a = F / m = 1 / 2 = 0.5 m/s 2 Answer: (A) During the first stage of the motion (on the rough surface), the forces on the box is balanced according to Newton s First Law since its speed is constant. Therefore the friction force equals the applied force. During the second stage of the motion (on the smooth surface), the forces on the box not balanced since the friction force is zero, only applied force is present. By Newton s Second Law, F = ma, since resultant force is not zero, acceleration is also not zero. Thus the box accelerates.
19 Answer: (A) By Newton s Third Law (action = reaction), both cart X and cart Y will experience the same amount of push force by the spring, as illustrated in the diagram below. For cart X: By Newton s Second Law, F = ma, the force from the spring, F s = 9 x m (1) Force cart Y: F s = 3m x a (2) From (1) & (2) 9 x m = 3m x a a = 9/3 = 3 m/s 2
20 Answers: Method 1 The block is moving in the same direction of F above (since friction force acts in the opposite direction of the motion). By Newton s Second Law, F = ma, the magnitude of the resultant force required to cause a deceleration of 2.0 m/s 2, Resultant force = 5 x 2.0 = 10 N To cause a deceleration, the resultant force must act in the direction opposite to the motion. Hence the friction force is 10 N greater that the applied force (20N). Thus friction force = = 30 N. Method 2 By Newton s Second Law, F = ma, resultant force, F is a vector quantity. Acceleration, a is also a vector quantity. The direction of F is the same as a. When an object decelerates the change in velocity is from a larger magnitude to a smaller magnitude, thus give rise to a negative change in velocity. Hence a deceleration is expressed a negative quantity of an acceleration.
21 From the free body diagram above, resultant force = F A - friction force (1) By Newton s Second Law, F = ma, where a = - 2 m/s 2 (deceleration), resultant force, F = 5 X (-2) = -10 N (2) From (1) & (2) F A - friction force = -10 N Friction force = F A + 10 = 30 N, where F A = 20 N 1 Answer: (B)
22 Free body diagram: By Newton s Third Law (action = reaction), the force, F C that pulls the carriage equals the force that the carriage is pulling the horse. In order for the horse to pull the carriage, the force provided by the horse, F H must be greater that F C.
23 Answers: Step 1: identify the forces involved. They are weights of the cradle and the builder, tension forces in each end of the rope. Step 2: Apply Newton s Third Law to i) the pulley and the cradle; and ii) pulley and the builder. By Newton s Third Law (action = reaction), the tension force in the rope hanging the cradle (T) equals tension force in the rope from the pulley (T). Tension force at both ends of the rope is also the same.
24 (a) Figure A Figure B Figure C Apply Newton s Third Law between the builder and the cradle:
25 The downward force exerted by the builder s feet on the cradle (refer to figure B) equals the upward normal force exert by the cradle to the builder (refer to figure C). From figure C, the forces action on the builder: normal reaction on the builder, the builder s weight and the upward force from the rope on the builder. The net upward force on the builder = Normal reaction the builder s weight (750 N) + the tension force from the rope, T (650 N). (b) Refer to figure B, The net force on the cradle = Tension (650 N) downward force exerted by the builder s feet on the cradle weight of the cradle (350 N) (c) Refer to figure A, the resultant force = 2 x 650 N 750 N 350 N = 200 N By Newton s Second Law, F = ma, where F = 200 N, m = a = F / m = 200 / 110 = 1.82 m/s 2 From (b), the net force on cradle = downward force exerted by the builder s feet on the cradle 650. Apply Newton s Second Law, F = ma: downward force exerted by the builder s feet on the cradle 350 = 35 kg x 1.82 m/s 2 Downward force exerted by the builder s feet on the cradle = x = N
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