0.30 kg. v = s/t s = vt = (30 ms -1 )(10 s) s = 300 m (s.f. may be 1, 2 or 3 since anything before decimal point is insignificant)

Transcription

1 Answers to orksheet ame: Date: Class: 3G1 Answer the following questions on a piece of foolscap. 1. Two forces of 15.0 and 20.0 are applied to a block of mass 0.30 kg in the directions shown in Fig 1 below. Assuming there is no friction present. hat is the acceleration of the block? kg 15.0 Taking left as negative (-ve), By ewton s Second Law of Motion, F R = = ma 5.0 = (0.30 kg) a a = 17 ms -2 ( 2 s.f.) 2. A car is travelling at 30 ms -1 on a level road. At this speed, the car has overcome a total force of 600 opposng the car. a. Calculate the distance travelled by the car in 10 s. v = s/t s = vt = (30 ms -1 )(10 s) s = 300 m (s.f. may be 1, 2 or 3 since anything before decimal point is insignificant) b. State the value of the driving force produced by the engine for a steady speed of 30 ms (constant velocity, hence no net forces, therefore driving force must be balanced by opposing forces) c. Explain why the car slows down when it climbs a hill even though the driving force is unchanged. This is because the ET FORCE acting on the vehicle is in the OPPOSITE direction compared to that of the velocity of the vehicle. This results in the velocity decreasing. (please turn to back for detailed explanation) 1

2 Driving Force, F D = 600 Y = mg Cosθ = mg X = mg Sinθ For those who are more adventurous, please look at the diagram above. The box represents the vehicle on the slope. The weight,, may be resolved into its two components (the dotted arrows) along an x-axis and y-axis chosen by you. The above x- and y-axes were chosen because they are convenient to work with. Once the is resovled into its components, it is completely replaced by them. It need not be considered anymore. It is as though there are now four forces acting on the vehicle (ie., F D, X and Y ; X may be shifted up and drawn on the vehicle), instead of the original three (ie., F D, ). Because the vehicle does not move up or down perpendicular to the slope, e may therefore infer that Y and cancels each other out. Hence, we are left with X and F D. If the vehicle slows down, there must be a net force acting downhill. e may hence infer that X > F D. Still cannot understand? See attachment at the end of this answer key. d. hile travelling at 30 ms -1 on the level road, the driving force becomes zero. The mass of the car is 800 kg. Calculate the deceleration of the car. F = ma ( 600 ) = (800 kg) a a = 0.75 ms -2 (s.f. not critical) 3. A box of mass 2.0 kg is pushed along a rough surface by an external force F of 10 such that it moves at constant speed. a. Comment on the following statement: Friction is both a friend and a foe. Friend Required for braking, required for walking so that you may not slip. Foe Reduces engine efficiency because energy is converted to heat which cannot be used. 2

3 b. Draw a free-body diagram of the 2.0 kg box, indicating its nature, line of action and magnitude. F F f c. If F is increased to 20.0, calculate the resulting acceleration. F R = 20.0 = ma = (2.0 kg) a a = 5.0 ms -2 (2 s.f.) 4. A 3000 kg truck is connected to a 1000 kg trailer by a tow-bar. The whole system moves at an acceleration of 2.0 ms -2. The friction acting on the truck is 800, and that on the trailer is 200. F D m truck = 3000kg Tow-bar m trailer = 1000kg F friction = 800 F friction = 200 a. Calculate the driving force exerted by the engine of the truck. Treating the whole system as a single body, F R = F D Ftotal friction = ma F D ( ) = (3000 kg kg)(2.0 ms -2 ) b. Calculate the tension acting on the tow-bar by the trailer. On trailer only, F R = T F f = ma T = ma + F f T = (1000 kg)(2.0 ms -2 ) T =

4 c. After some time, the whole system travels at a constant velocity of 10ms -1. State the new value of the driving force (Constant velocity, ie. o net force acting on the whole system, hence driving force must be equal to the frictional forces). d. Describe the motion of the system when the engine is suddenly cut off in terms of its velocity and acceleration. The system continues to move forward with a decreasing velocity until it comes to rest. The acceleration is throughout this motion is 0.25 ms A 1200 kg lift accelerates uniformly from a velocity of 50 ms -1 to 60 ms -1 in 5 seconds. a. Draw a free-body diagram representing the forces acting on the lift If lift accelerating downwards, T If lift accelerating upwards, T. b. Calculate the acceleration of the lift. a = (v-u)/t a = (60 50) / 5 a = 2 ms -2 c. Calculate the tension in the cable which holds the lift. T = ma T = ma + You may omit intermediate steps. I only wrote T = ma + mg these steps for those who cannot follow the math. T = m (a + g) T = 1200 kg ( 2 ms ms -2 ) T =14400 d. A man weighing 500 enters the lift. Determine the magnitude of the force acting on his feet. The normal contact force of the lift acting on the man is 500, if the lift is not accelerating. However, if the lift is accelerating upwards at 2 ms -2, then, On the man, = ma = ma + mg = 50kg (2 ms -2 ) = 600 4

5 The normal contact force acting on his feet is 600. Likewise, he will also exert a force of 600 back on the lift. Hence, if you place a weighing scale in the lift, and have him stand on it, the scale will read 600 as well, because the weighing scale measures the normal contact force of the man on the scale. 6. By means of a scale drawing, determine the resultant force acting on the block below: 40.0 This diagram is not drawn to scale, however, if you had drawn to scale, you should be able to obtain 40.0 of the resultant force. ~end~ 5

6 Supplementary otes m = 10.0kg a) Determine the acceleration of the 10.0 kg box above down the frictionless slope. m = 10.0kg the slope. Y X By rearranging the free-body diagram a little shifting the eight,, so that it may originate from the same point as the ormal Contact Force,, we obtain the diagram on the left. Let us resolve the eight,, into its X- and Y-components. Recall from vector addition that Resultant Vectors are summed from Component Vectors. e may also reverse the process by resolving the Resultant Vector into its components. Let X be the X-component of the weight that is perpendicular to the surface of the slope and Y be the Y-component of the weight that is parallel to the surface of The Y-component of the eight, Y, must be equal in magnitude as the ormal Contact Force. If it is not, the box will sink perpendicular to the slope surface into the slope, or fly up! Since they are equal in magnitude, but pointing in opposite directions, they cancel each other out. The only force that needs to be dealt with is the X-component of the eight, X. (ote that the eight no longer is of concern because it has been resolved into its components; ie. It has been replaced by the components). X is the only force left. In fact, it is the vector sum of and. It is thus the resultant force in this example. Hence, But XX = sin, Sub = mmmm FF RR = XX = mmaa sin = mmmm mmmm sin = mmmm aa = mmmm sin mm = gg sin = 5.0 ms -2 6