NCSP 2 nd Ed Chapter 4 Solutions Forces

Transcription

1 NCSP 2 nd Ed Chapter 4 Solutions Forces Question 1 (a) Answer 2.5 N north (b) Answer 101 N north (c) Answer 32 N, S72 W or W18 S (d) Answer 16 N, N22 W or W 68 S NCSP Chapter 4 Solutions 1

2 Question 2 p77 Answer. (a) Forces are balanced as the force vectors are the same length. (b) The forces on the book would not be balanced. The only force would be the weight pointing down, so the book would accelerate downwards. Question 3 p78 Answer. (a) When pulled slowly the thread A would have to support weight of the ball and the pulling force of the string so string A would break. (b) when pulled quickly the inertia of the ball would resist a change in motion so the bottom string would break. Question 5 p78 (a) the water keeps moving in a straight line when your hand is pulled away so the water is flicked off. (b) the water keeps moving in a straight line when the edge of the umbrella edge keeps turning towards the centre so the water is flicked off. (c) while riding you and your bike tend to continue in a straight line so you don t fall over. When stationary, there is no force keeping the bike upright. (d) falling off a building is like jumping into a swimming pool except that the landing is done over a shorter time when falling off a building. The rapid deceleration is what kills you. Q6 p80 Question 7 p82 (a) a = (v-u)/t = (25-0)/8.5 = 2.94 ms -1 ; F = ma = 1000 x 2.94 = 2940 N. (b) a = F/m = 25/15 = 4.6 ms -2 ; v = u + at = x 2 = 9.2 ms -1. (c) m = F/a = 1000/10 = 100 kg; a = (v-u)/t hence t = (v-u)/a = (40-10)/10 = 3 s. (d) a = (v-u)/t = ( )/ (1.5 x 60) = 0.25/90 = ms -2 ; F = ma = 200/1000 kg x = 5.5 x 10-4 N. (e) m = F/a = 150/(-2.2) = kg; v = u + at or u = v at = 0 (-2.2) x 4 = 8.8 ms -1 Question 8 u = 30 ms -1, v = 0 ms -1, s = 100 m, m = 2000 kg. Calc F v 2 = u 2 2as; 0 2 = x a x 100; a = (0-900)/(2x100) = -4.5 ms -2 F = ma = 2000 x -4.5 = 9000 N Question 9 u = 173 km/h = 173 x1000/3600 = 48 ms -1 ; s = 0.66 m; m = 55 kg. Calc F NCSP Chapter 4 Solutions 2

3 Acceleration (m/s^2) v 2 = u 2 + 2as; 0 2 = x a x 0.66; a = -2304/(2 x 0.66) = ms -2. F = ma = 1745 x 55 = N (or N to 2 significant figures) decelerating force. Question 10 (a) False. As a = (v-u)/t you get the same acceleration for both situations, thus F will be the same too (F = ma). (b) True. For rest to 60 km/h over 100m: v 2 = u 2 2as; (60 x 1000/3600) 2 = 0 2 = 2 x a x 100; a = 277.8/(2 x a x 100); hence a = 1.39 ms -2. For 200m: a = 277.8/(2 x a x 200) = 0.69 ms -2. This half the acceleration so the force will be half as F = ma. (c) True. However, it may be moving in a direction opposite to the net force but will always accelerate in the direction of the net force. For example, a ball thrown upwards will be travelling in the upwards (positive) direction even through the net force (gravity) is down (negative direction) and the acceleration is directed downwards (negative direction). It will move upwards (positive direction) but will slow down until it reaches zero velocity at maximum height and then will start to accelerate and speed up downwards. (d) False. The force is proportional to the acceleration, and as the acceleration is equal to the change in speed divided by the time elapsed, the time is important. Question 11 Graph suggests acceleration is proportional to the force. Graph of Force vs acceleration y = x R² = Force (N) Question 12 (a) Hanging mass Mass of glider and m2 Accel Applied F m1 (g) (g) (ms -2 ) (N) NCSP Chapter 4 Solutions 3

4 (b) 12 (c) Force is proportional to acceleration (linear) 12 (d) If friction was present the net force would be less and the acceleration would be less. 12 (e) The hanging mass would be kept constant and thus provide a constant force. The mass added to the glider could be changed. Question 13. (a) the racquet is hit by the ball; (b) the road pushes back on the horse; (c) the earth pushes back against the horse s feet; (c) the ground pushes back against the Click Beetle s feet; (d) as the Earth pulls the man towards the ground the reaction is that the man pulls the Earth upwards towards him. Question 14. (a) FN = mg = 2.5 x 10 = 25 N (upwards direction); (b) A large mass of 500kg would have a downward force of 5000 N due to gravity. The normal reaction would have to be a 5000 N force upwards. It is unlikely that a table could provide this and the legs may break. Question 15. (a) FW = mg = m x 10 N on Earth. On Saturn FW = m x 10.5 N. Hence, on Saturn the weight would be 10.5m/10m = 1.05 times as great; (b) The Earth is not a perfect sphere, and the distribution of mass is also not even. NCSP Chapter 4 Solutions 4

5 Question 16. (a) A mass of 7 lb 8 oz = 7.5 lb = 7.5/2.2 = 3.4 kg. FW = mg = 3.4 x 10 = 34 N. Congratulations to the mother and father; (b) you could use a Wig-Wag machine that measure the rate of oscillation of a metal bar (see Figure 4.10, page 78). Question 17. Apparent weight is not considered a physics term but is used to show the weight of an object as measured on a spring balance. The weight of an object doesn t change on Earth as it is always equal to FW = mg (= m x 10) N. It is better to use the term Scale Reading on a balance calibrated in Newton. (a) when floating, a 70kg person has a mass of 70kg and a weight of 700N. Because there is an upward force supplied by the water and equal to the weight, the apparent weight or scale reading is 0 N; (b) When free-falling, the weight is still 700 N but scales under the body of the person or spring balance attached to the body would be also in free-fall so the apparent weight would be zero; (c) the balance would have to provide an upward force to support the person s weight (700 N) but also provide an accelerating force upward equal to m x a (= 70 x 0.5 = 35 N). Thus the scales would read 735 N and this would be the apparent weight; (d) when the lift accelerates downward the scales would have to provide enough force to change the downwards acceleration from 10 ms -2 to 0.5 ms -2. It would only have to provide a force of F = ma = 70 x 9.5 = 665 N. So this is the apparent weight. You could also think that 35 N of the weight was used to create the downwards acceleration of 0.5 ms (a) FN = FW cos = 30 x 10 x cos 35 = N; (b) As there is no friction FP = FW sin = N; a = F/m = 172.1/30 = 5.7 ms -2. Question 19 Fnet = 10 N = ma; a = 10/(2 + 3) = 2 m s -2 T = m2kg a = 2 x 2 = 4 N to the right Question 20 (a) Fnet = F1 + F2 = = 68.6 N a = Fnet/mtotal = 68.6/(15 + 8) = 2.98 m s -2 T1 = force to support 15 kg less the accelerating force (because it is acting in the same direction as the weight) = x 2.98 = N NCSP Chapter 4 Solutions 5

6 T2 = force to support 8 kg plus the accelerating force (because it is acting in the oppositee direction as the weight): (8 x 2.98) = N Note that T1 has to equal T2 (which it does). Alternative solution: For 15 kg block (m1) falling: FW1 T = Fnet T = 15a (a is positive as it is directed down as is the weight) T = a (Equation 1) For the 8 kg block moving up (a is negative): FW2 - T = 8 a T = -8a 78.4 T = -8a T = a (Equation 2) Solve simultaneous equations: Equation 1 = Equation 2 T = a = a 68.6 = 23 a a = 2.98 m s -2 T = a = = N (b) Fnet = F1 + F2 = mg + 0 (the weight of the 0.8 kg mass does not pull on the string) = = 9.8 N a = Fnet/mtotal = 9.9/1.8 = 5.44 m s -2 T1 = m1g = m1a = = 4.45 N T2 = 0 + m2a = = 4.45 N (as expected T1 has to equal T2) Alternative solution: For 1.0 kg block (m1) falling: FW1 T = Fnet T = 1.0 a (a is positive as it is directed down as is the weight) T = a (Equation 1) For the 0.8 kg block sliding horizontally: T = m2a = 0.8 a (the weight has no effect in the horizontal direction and is in the same direction as the tension (T); we also assume friction is zero. T = 0.8a (Equation 2) Solve simultaneous equations: Equation 1 = Equation 2 NCSP Chapter 4 Solutions 6

7 T = a = 0.8a 9.8 = 1.8 a a = 5.44 m s -2 T = 0.8 x 5.44 = 4.45 N Question 21 p99 FA = 220 N Ff = 220 N FW = mg = = 392 N FN = FW = 392 n = Ff/FN = 220/392 = 0.56 Question 22 p99 FW = mg = = N [Note: 1 tonne = 1000 kg] FN = FW = N Ff = FN = = N FH = Ff = N Question 23 p 100 (a) FA = 350 N FH = FA cos = 350 cos 38 = N Ff = N (b) FW = mg = = 784 N FV = FA sin = 350 sin 38 = N FN = FW FV = = N (c) = Ff/FN = 275.8/568.5 = Question 24 Ff = FN FA cos = FN FA cos 40 = (FW + FA sin ) FA = 0.12 ( FA sin 40 ) FA = FA FA = NCSP Chapter 4 Solutions 7

8 FA = N Question 27 Fnet = Fdown - Fup ma = mg sin - µ mg cos a = g sin - µ g cos a = 9.8 sin cos13.8 a = a = 4.61 m s -2 Kinematics formula: u 2 = 2as u 2 = u = m s -1 Question 31 p 104 (a) ( )N - 18S = 25 N north; (b) ( )U - 6.5D = 10.1 N up; (c) By Pythagoras Theorem: FR = ( ) = 36.4 N; = tan -1 (10/35) = W16 S, or S74 W. (d) Pythagoras: ( ) = 1.8 N; = tan -1 (1.5/1) = W56 N. NCSP Chapter 4 Solutions 8

9 Question 32 p 104 Volume = l b d = 25 x = m 3 m = density V = = kg (202.5 g) Question 33 p104 (a) 1 st (you want to continue at your current speed but the car accelerates and presses on your back); (b) 3 rd (the water spurting out causes an equal and opposite reaction); (c) 2 nd (the force acting on it is constant and so is its mass, so by Newton s 2 nd law: a = F/m = constant; (d) 1st (the object was in a constant state of motion (at rest) but when acted on by an unbalanced force it accelerates in the direction of the unbalanced force which is down). Question 35 (a) F = ma = = N east; (b) a = (v -u)/t = (2-0)/3 = 0.67 m s -2 ; F = ma = = N (c) m = F/g = 25/9.8 = 2.55 kg; F = ma = = 25 N (d) v 2 = u as; 5 2 = a 4.0; a = 25/8 = m s -2 ; F = ma = = N (e) a = (v u)/t = (60 40)/( ) = 4000 m s -2 ; F = ma = = N Question 41 (a) 3rd law Throw your jacket off behind you to accelerate; (b) 2nd law The faster you throw, the greater the acceleration; (c) 1st law To change direction, throw something to the side Question 42 Rolling and sliding friction is independent of area. Starting (static) friction varies with area. Question 43 High friction is more important. The rubber compounds in the tyres are selected to perform within a specific temperature range. When too cold, less grip means spins or having to travel slower on corners. Travelling slower is a problem for grip as the car aerodynamics are such that they can only effectively corner at higher speed (pressed onto the tarmac). Question 44 (a) NCSP Chapter 4 Solutions 9

10 (b) F = ma; m = F/a = 1/gradient = 1/0.3 = 3.3 kg (c) Line would cut x axis at somewhere greater than zero. Figure For question 76. Question 45 (a) mass added (kg) mass of trolley (kg) total mass (kg) t (s) a= 2s/t 2 (m s -2 ) F = ma (N) (b) 0.25 m s 2, 0.18 m s 2, 0.15 m s 2 ; NCSP Chapter 4 Solutions 10

11 (c) (d) m 1/a; (e) N in each case; (f) (i) about m s 2 from the graph; a = F/m = 0.175/1.3 = m s -2 (ii) s = ½ at 2 ; t 2 = 2s/a = 1/0.134 = 7.46; t = 2.73 s (g) The force from the rubber bands would have to be constant. Question 46 (a) combustion 60%, engine 21.4%, transmission 1.6%, accessories 2.2%, tyres 5.2%, air 4.9%, brakes 4.7%; (b) decrease i, ii, v, vi, vii; increase iii, iv Question 47 Steel plate drags along ground. Loss of control; dig up road Question 52 s = 290 m; = 0.60 Ff = FN = m mg Ff = ma a = Ff/m = mg/m = g = = 6 ms -2 v 2 = u 2 +2as 0 = u u = 59 ms -1 (212 km h -1 ) Question 53 Basic relationships: FH = Ff = FAcos ; FN = FW FA sin Ff = FN Ff = (FW FAsin ) FAcos = (FW - FAsin ) FA 0.74 = 0.1(750 FA 0.67) FA = 75 FA = 93 N NCSP Chapter 4 Solutions 11

12 Question 54 u = 20 m s -1 ; F = ; s = 22.0 m a = F/m = 6 x 10 3 /1500 = 4.0 ms -2 v 2 = u 2 + 2as = v 2 = v = 15 m s -1 Question 55 Convert 60 km h -1 to m s -1 = 16.7 m s -1 At terminal velocity FW = ½ C A v 2 ½ C A1 v1 2 = ½ C A2 v2 2 cancelling ½ C from both sides A1 v1 2 = A2 v2 2 A1 v1 2 = 2A1 v2 2 (replacing A2 with 2A1) v1 2 = 2 v = 2 v = 2 v2 2 v2 = 11.7 ms -1 Question 56 u = m s 1 ; FUP = N; FW = N FNET = FUP FW FNET = N N = N a = F/m = / = m s 2 t = s/v = / = seconds s (up) = ut + ½ at 2 = 0 + ½ ( ) 2 = m (15.3 cm) Question km h -1 = 30000/3600 = 8.33 m s -1 a1 = µg = = m s -2 a2 = µg = = m s -2 STAGE 2: v 2 = vmid 2 + 2as = vmid x = vmid 2 vmid = 10.4 m s -1 STAGE 1: vmid 2 = u 2 + 2as = u NCSP Chapter 4 Solutions 12

13 u 2 = u = 15.2 m s -1 NCSP Chapter 4 Solutions 13