140 iii. Calculate the car deceleration. = + 0 = 80 + 25 = − 80 25 = −3.2 m/s2 iv. Calculate the car deceleration. 1 = 1 2 × (15 + 80) × 18.57 1 = 882.14 2 = 35 × 80 2 = 2800 3 = 1 2 × 80 × 25 3 = 1000 = 1 + 2 + 3 = 882.14 + 2800 + 1000 = 4682.14
141 16. The movement of a particle is expressed in relation of ( = 5 3 − 4 2 + 3 + 12) m, where x and t are respectively in meters and seconds. If t = 8 seconds, determine i. Position. ii. Velocity. iii. Acceleration. Answer: i. Position. = 5 3 − 4 2 + 3 + 12 = 5(8) 3 − 4(8) 2 + 3(8) + 12 = 2560 − 256 + 24 + 12 = 2340 ii. Velocity. = 5 3 − 4 2 + 3 + 12 = = 15 2 − 8 + 3 when = 8 = 15(8) 2 − 8(8) + 3 = 960 − 64 + 3 = 899 / iii. Acceleration. = 15 2 − 8 + 3 = = 30 − 8 when = 8 = 30 − 8 = 30(8) − 8 = 232 m/s2
142 17. A ball is thrown downward from a 60m tower with an initial velocity of 20m/s, Determine the velocity at which it hits the ground and the time of travel. Answer: 2 = 2 + 2 2 = 202 + (2)(9.81)(60) 2 = 1577.2 = 39.71 / The time of travel = + 39.71 = 20 + (9.81) = 39.71−20 9.81 = 2
143 18. A car travels in a straight line with initial velocity 10m/s and it accelerates uniformly at 2 m/s2 for 300m. Then, it travels at a constant velocity for 20 seconds. Finally, it slows at a constant deceleration for 15 seconds until it stops. i. Sketch a velocity-time graph. ii. Calculate the constant velocity of the car. iii. Calculate the deceleration of the car. iv. Calculate the total distance for the journey. Answer: i. Sketch a velocity-time graph. ii. Calculate the constant velocity of the car. 2 = 2 + 2 2 = 102 + (2)(2)(300) 2 = 1300 = √1300 = 36.06 / iii. Calculate the deceleration of the car. = + 0 = 36.06 + 15 = − 36.06 15 = −2.404 m/s2
144 iv. Calculate the total distance for the journey. 1 = 300 2 = 20 × 36.06 = 721.1 m 3 = 1 2 × 15 × 36.06 = 270.45 = 1 + 2 + 3 = 300 + 721.1 + 270.45 = 1291.55
145 19. A car has an acceleration and a deceleration of 6 m/s. If it starts from rest and can have a maximum velocity of 60m/s, predict the shortest time it can travel for 1200m before it stops. Answer: Time to reach. = + 60 = 0 + 6 = 60 6 = 10 Distance to reach 60 m/s = + 1 2 2 = 0 + 1 2 (6)(10) 2 = 300 Total distance for acceleration and a deceleration 300 x 2 = 600 m V (m/s) 60 t(s) Distance travel at constant for velocity 60 m/s 1200 – 600 = 600 m Time travel at constant velocity 60 m/s: 600 ÷ 60 =10 sec
146 Total time, t =10 + 10 + 10 = 32 s
147 20. The coordinate of a car which is confined to move along a straight line is given by equation, = 2 3 − 24 + 6 where s is the distance travelled by the car measured in meter from an origin and t is the duration of travel in seconds. Calculate: i. The time required for the car to reach the velocity of 72 m/s from its initial condition at = 0. ii. The acceleration of the car when the velocity is 30 m/s. iii. The net displacement of the car during the interval from = 1 to = 4 . Answer: i. The time required for the car to reach the velocity of 72 m/s from its initial condition at = 0. = 2 3 − 24 + 6 = = 6 2 − 24 6 2 − 24 = 72 2 = 72+24 6 2 = 16 = √16 = 4 ii. The acceleration of the car when the velocity is 30 m/s. = 6 2 − 24 If = 30 / = 6 2 − 24 6 2 − 24 = 30 2 = 30+24 6 2 = 9 = √9 = 9
148 = = 12 = 12 = 12(3) = 36 m/s2 iii. The net displacement of the car during the interval from = 1 to = 4 . = 2 3 − 24 + 6 = 1 = 4 ∆ = 4 − 1 = [2(4) 3 − 24(4) + 6] − [2(1) 3 − 24(1) + 6] = 38 − (−16) = 54
149 21. A ball is thrown vertically upward as shown in figure below with a speed of 15 m/s. Predict the time when it returns to its original position. Answer: = 15 / , = 0, = −9.81 m/s2 = + 0 = 15 − 9.81 = 15 9.81 = 1.529 = = 1.529 ∴ = 1.529 + 1.529 = 3.058
150 22. A shaft starts from rest and reaches velocity 250 rpm in 10 seconds. Calculate the angular acceleration of the shaft and the number of rotation within the time. Answer: 1 = 0 2 = 2 60 2 = 2(250) 60 2 = 26.18 / = 2 − 1 = 26.18 − 0 10 = 2.618 rad/s2 Number of rotations: 2 2 = 1 2 + 2 (26.18) 2 = 0 + (2)(2.618) 685.3924 = 5.236 = 685.3924 5.236 = 130.9 Number of rotations =130.9 2 = 20.83
151 23. An object being thrown up from the top of 100 meters building from the ground with a velocity 20 m/s. Determine: i. The velocity of an object exactly when it hits the ground. ii. Time taken from it starts until it reaches the ground. Answer: i. The velocity of an object exactly when it hits the ground. 2 = 2 − 2 0 2 = (20) 2 − (2)(9.81) 0 = 400 − 19.62 = 400 19.62 = 20.39 2 = 2 + 2 2 = 0 + (2)(9.81)(100 + 20.39) 2 = 2362.0518 = √2362.0518 = 48.6 /
152 ii. Time taken from it starts until it reaches the ground. moving up = + 0 = 20 − (9.81) = 20 9.81 = 2.04 moving down = + 48.6 = 0 + 9.81 = 48.6 9.81 = 4.95 , = 2.04 + 4.95 = 6.99
153 24. A car starts from rest with a constant acceleration of 1.25 m/s2 until it achieves 50 km/h. The velocity maintained as far as 1.2 km. When the brake is applied it stops with 10 second with constant deceleration. i. Calculate the total time taken during the journey. ii. Determine the total distance during the journey. Answer: i. Calculate the total time taken during the journey. 50 /ℎ = 50 ℎ × 1000 1 × 1 ℎ 3600 = 13.89 / 1, = + 13.89 = 0 + 1.25 = 13.89 1.25 = 11.112 2, = + 1 2 2 1200 = 13.89
154 = 1200 13.89 = 86.39 3 = 10 , = 1 + 2 + 3 = 11.112 + 86.39 + 10 = 107.5 ii. Determine the total distance during the journey. 1 = 1 2 × 13.89 × 11.112 = 77.17 2 = 13.89 × 86.39 = 1199.96 3 = 1 2 × 13.89 × 10 = 69.45 , = 1 + 2 + 3 = 77.17 + 1199.96 + 69.45 = 1346.58
155 25. A car is travelling along a straight road at 30 km/h and the speed increases to 100 km/h in 20 s. Determine the distance travelled. The answer must be in SI unit. Answer: = 30 ℎ × 1000 1 × 1 ℎ 3600 = 8.33 / = 100 ℎ × 1000 1 × 1 ℎ 3600 = 27.78 / = 1 2 ( + ) = 1 2 (27.78 + 8.33)20 = 361.1
156 26. A car starts from rest and accelerates uniformly for 50 seconds and reaches a velocity of 70 m/s at the end of the acceleration. Its velocity is maintained for a while and then it stops within 60 seconds with 60 seconds with constant deceleration. The total distance travelled by the car is 11.5 km. i. Sketch a velocity-time graph. ii. Calculate the acceleration of the car. iii. Calculate the time taken for the journey. Answer: i. Sketch a velocity-time graph. ii. Calculate the acceleration of the car. = + 70 = 0 + 50 = 70 50 = 1.4 m/s2 iii. Calculate the time taken for the journey. 1 = 1 2 × 70 × 50 = 1750 2 = 70 × = 70
157 3 = 1 2 × 70 ×60 = 2100 , = 1 + 2 + 3 11500 = 1750 + 70 + 2100 11500 = 3850 + 70 70 = 11500 − 3850 70 = 7650 = 7650 70 = 109.3 ∴ time taken = 50 + 60 + 109.3 = 219.3
158 27. A vehicle moves in a straight road with a displacement defined by = (0.6 3 + 0.3 2 ) where t is in second and s is in meter. When = 6 , identify: i. Velocity. ii. Acceleration. iii. Predict whether the vehicle is in the motion of accelerating or decelerating within the duration time of {1 ≪ ≪ 6} . Answer: i. Velocity. = = 1.8 2 + 0.6 When = 6 = 1.8(6) 2 + 0.6(6) = 64.8 + 3.6 = 68.4 / ii. Acceleration. = = 3.6 + 0.6 When = 6 = 3.6(6) + 0.6 = 22.2 m/s2 iii. Predict whether the vehicle is in the motion of accelerating or decelerating within the duration time of {1 ≪ ≪ 6} . = 1.8 2 + 0.6 When = 1 = 1.8(1) 2 + 0.6(1) = 1.8 + 0.6 = 2.4 / By comparing the velocity from = 1 to = 6 , shows that vehicle is in acceleration by the increasing its velocity.
159 28. A vehicle moves in a straight line such that for a short time its velocity is defined by = 0.9 2 + 0.6 m/s where is in second. When = 3 , determine: i. Displacement. ii. Acceleration. Answer: i. Displacement. = ∫ (0.9 2 + 0.6) 3 0 = [0.3 3 + 0.3 2 ] 3 0 = [0.3(3) 3 + 0.3(3) 2 ] − 0 = 8.1 + 2.7 = 10.8 ii. Acceleration. = 0.9 2 + 0.6 = = 1.8 + 0.6 When = 3 = 1.8(3) + 0.6 = 6 m/s2
160 29. A car starts from rest and accelerates uniformly for 70 seconds and reaches a velocity of 80 m/s at the end of the acceleration. Its velocity is maintained for a while and then it stops within 65 seconds with constant deceleration. The total distance travelled by the car is 12.2 km. i. Draw a velocity-time graph. ii. Determine the acceleration of the car. iii. Calculate the time taken for the journey. iv. Determine the deceleration of the car. Answer: i. Draw a velocity-time graph. ii. Determine the acceleration of the car. = + 80 = 0 + 70 = 80 70 = 1.143 m/s2
161 iii. Calculate the time taken for the journey. 1 = 1 2 × 70 × 80 = 2800 2 = 80 × = 80 3 = 1 2 × 80 × 65 = 2600 , = 1 + 2 + 3 12200 = 2800 + 80 + 2600 12200 = 5400 + 80 80 = 12200 − 5400 80 = 6800 = 6800 80 = 85 ∴ time taken = 85 + 70 + 65 = 220 iv. Determine the deceleration of the car. = + 0 = 80 + 65 = − 80 65 = −1.231 m/s2
162 30. A train starts from rest at a station with constant acceleration of 2.5 m/s2 until it achieves velocity of 70 km/h. Then, the train decelerate until it stop in 10 s. Determine: i. Distance travelled by the train. ii. Deceleration of the train. Answer: i. Distance travelled by the train. = 70 ℎ × 1000 1 × 1 ℎ 3600 = 19.44 / v (m/s) 19.44 1 2 = + 19.44 = 0 + 2.5 = 19.44 2.5 = 7.78 1 = 1 2 × 19.44 × 7.78 = 75.62 2 = 1 2 × 19.44 × 10 = 97.2 t (s)
163 , = 1 + 2 = 75.62 + 97.2 = 172.82 ii. Deceleration of the train. = + 0 = 19.44 + 10 = −19.44 10 = −1.944 m/s2
164 CHAPTER 4 KINETICS OF PARTICLES 1. State the following terms: i. Kinetic ii. Work Answer: i. Kinetic: Relationship between the change in motion of a body and the forces that cause this change. ii. Work: When the particle undergoes a displacement in the direction of the force. 2. Define Newton’s second law of motion and give ONE (1) example in real life phenomenon. Answer: A particle acted upon by an unbalanced force experiences an acceleration a that has the same direction as the force and a magnitude directly proportional to the force. = . Example: Pushing a car is much easier than pushing a truck because the car has less mass than the truck. or Pushing an empty shopping cart is easier than pushing a loaded shopping cart. (accept any appropriate example)
165 3. The principles of relationship between work and energy can be used to solve some kinematics variables in kinetic related problems. i. Explain the relationship between work and energy. ii. Reprsent the relationship between work and energy in mathematical equantions. ∑− = 1 2 2 − 1 2 2 Answer • Work done by a force on body is the force times the displacement of the body in the direction of the force. • Energy is the ability to do work. • The amount of energy something has is the amount of work it can do. Reprsent the relationship between work and energy in mathematical equantions. ∑− = 1 2 2 − 1 2 2 4. Define the following terms: i. Potential Energy ii. Kinetic Energy Answer i. Potential Energy • Exists when an object or rigid body is at a certain height or due to the position of the object. • Can be calculated using the formula = ℎ ii. Kinetic Energy Exist when an object or rigid body is in motion where the motion can be horizontal, vertical or in any direction. Can be calculated using the formula = 1 2 2
166 5. State the Newton’s Second Law and give TWO (2) examples of its application Answer: Newton's Second Law states that when an unbalanced force acts on a particle, the particle will accelerate in the direction of the force with a magnitude that is proportion to the force. Example of its application: pushing a cart, kicking a ball 6. An object of mass 5 kg is dropped from a height of 35 m from a building. Based on the situation below, express the value of: i. The potential energy possessed by the object before it fell. ii. The kinetic energy possessed by the object before it fell. iii. The potential energy possessed by the object after it fell and touched the ground. iv. The kinetic energy possessed by the object after it fell and touched the ground. Answer: i. The potential energy possessed by the object before it fell. = ℎ = 5 × 9.81 × 35 = 1716.75 J ii. The kinetic energy possessed by the object before it fell. = 1 2 2 = 1 2 (5)(0) = 0 iii. The potential energy possessed by the object after it fell and touched the ground. = ℎ = 5 × 9.81 × 0 = 0
167 iv. The kinetic energy possessed by the object after it fell and touched the ground. 2 = 2 + 2 = 0 + 2(9.81) (35) = 686.7 m/s = 1 2 2 = 1 2 (5)(686.7) = 1716.75 J 7. The 75 kg crate shown in figure below rest on a horizontal surface for which the coefficient of kinetic friction is = 0.3. If the crate is subjected to a 600 N towing force as below: i. Draw a Free Body Diagram (FBD). ii. Calculate the velocity of the crate after 5 second starting from rest. Answer: i. Draw a Free Body Diagram (FBD).
168 ii. Calculate the velocity of the crate after 5 second starting from rest. + ↑ ∑ =0 − + 600 30 = 0 − (75)(9.81) + 300 = 0 − 735.75 + 300 = 0 − 435.75 = 0 = 435.75 = = (0.3)(435.75) = 130.725 N + → ∑ = 30 − = (600) 30 − 130.725 = 75 519.62 − 130.725 = 75 388.895 = 75 = 388.895 75 =5.19 m/s2 = + = 0 + (5.19)(5) = 25.95 /
169 8. Figure below shows a man is pushing a machine on a rough surface. Given that the pushing force 600 N, acted 48° from the negative x-axis. Total mass of the machine is 20 kg. If the kinetic friction, 0.4 ∶ i. Draw the free body diagram of forces that acted on the machine. ii. Calculate the normal force and friction force. iii. Calculate the acceleration of the machine. Answer: i. Draw the free body diagram of forces that acted on the machine. ii. Calculate the normal force and friction force. + ↑ ∑ =0 −600 48° − 196.2 + = 0 − 445.89 − 196.2 + = 0 −642.09 + = 0 = 642.09
170 = = (0.4)(642.09) = 256.836 N iii. Calculate the acceleration of the machine. + → ∑ = 600 48° − = 20 401.48 − 256.836 = 20 = 144.644 20 = 7.23 m/s2 9. A particle of 4 kg mass is being pulled across a smooth horizontal surface by a horizontal force. The force does 24 J of work in increasing the particle’s velocity from 5 m/s to v m/s. Express: i. The value of v after 15 seconds. ii. The value of the position for the particle after 15 seconds. Answer: i. The value of v after 15 seconds. 1−2 = 2 − 1 24 = 1 2 2 − 1 2 2 24 = 1 2 (4) 2 − 1 2 (4)5 2 24 = 2 2 − 50 74 = 2 2 74 2 = 2 = √37 v = 6.08 m/s
171 ii. The value of the position for the particle after 15 seconds. S = ( + 2 ) = ( 5 + 6.08 2 )15 = 83.1 m 10. The 60 kg crate in figure below rests on a horizontal surface for which the coefficient of kinetic friction = 0.3. If the crate is subjected to a 400 N towing force as shown, calculate: i. The value of normal force and the friction force. ii. The value of an acceleration of the crate. iii. The velocity of the crate after 3 seconds starting from rest. Answer: i. The value of normal force and the friction force. + ↑ ∑ =0 + 400 30° = = −400 30° + (60 × 9.81) = −200 + 588.6 = 388.6 = = 0.3 × 388.6 = 116.58 N
172 ii. The value of an acceleration of the crate. + → ∑ = 400 30° − = 60 400 30° − 116.58 = 60 60 = 229.83 = 229.83 60 = 3.83 m/s2 iii. The velocity of the crate after 3 seconds starting from rest. Given t = 3s, from rest u = 0 = + = 0 + (3.83)(3) = 11.49 m/s
173 11. A 1000 kg container rest on a horizontal surface which the coefficient of kinetic friction is = 0.3.The container is subjected to 6000N towing force acting at angle of 45 0 as shown. i. Express the system with the aid of free body diagram. ii. Express the value of the acceleration for the container in a mathematical method. Answer: i. Express the system with the aid of free body diagram.
174 ii. Express the value of the acceleration for the container in a mathematical method. + → ∑ = 6000 45° − = 1000 -------------------------------- equation 1 + ↑ ∑ =0 + 6000 45° = = −6000 45° + (1000 × 9.81) = −4241.64 + 9810 = 5567.36 = = 0.3 × 5567.36 = 1670.208 N Substitute = 1670.208 N into equation 1 6000 45° − 1670.208 = 1000 4242.64 − 1670.208 = 1000 2572.433 = 1000 = 2572.433 1000 a = 2.572 m/s2
175 12. A particle of 2 kg mass is being pulled across a smooth horizontal surface by a horizontal force. The force does 39 Joule of work in increasing the particle's velocity from 5ms-1 to v ms-1 . Calculate; i. The value of v. ii. The value of an acceleration for the particle. iii. The position of the particle after 15 seconds. Answer: i. The value of v. = 2 − 1 = 1 2 2 − 1 2 2 39 = 1 2 (2) 2 − 1 2 (2)(5) 2 39 = 2 − 25 2 = 39 + 25 2 = 64 = √64 = 8 / ii. The value of an acceleration for the particle after 15 seconds. = + 8 = 5 + 15 8 − 5 = 15 = 3 15 = 0.2 m/s2 iii. The position of the particle after 15 seconds. = + 1 2 2 = (5)(15) + 1 2 (0.2)(15) 2 = 75 + 22.5 = 97.5
176 13. A vehicle with a weight of 500 kg is driven downhill which is 6° incline and at a speed of 60 km/h as shown in Figure below. When the brakes are applied, it will cause a constant total braking force of 250N. Based on this information: i. Draw a free body diagram for this vehicle. ii. Calculate the kinetic energy for this vehicle. iii. Calculate the work done by the vehicle Answer i. Draw a free body diagram for this vehicle. ii. Calculate the work done by the vehicle. . = 60 /ℎ = 60 × 1000 3600 = 16.67 / = 1 2 2 = 1 2 (500)(16.67) 2 = 69472.225
177 iii. Calculate the work done by the vehicle. When the car stops, 2 = 0 Work done, = 2 − 1 = 0 − 69472.225 = −69472.225
178 14. A 1600 kg crate is pulled along the ground with a constant speed of a distance for 25 m, using a cable that makes a horizontal angle of 15°. Determine the tension in the cable. The coefficient of kinetic friction between the ground and the crate is = 0.55. Answer: + ↑ ∑ =0 + 15 = 15696 = 15696 − 15° ----------------------------- equation 1 + → ∑ = cos 15° − = 0 cos 15° − 0.55 = 0 ---------------------------------------------------equation 2 Substitude = 15696 − 15° into equation 2 cos 15° − 0.55(15696 − 15°) = 0 0.966 − 8632.8 + 0.142 = 0 1.108 = 8632.8 = 8632.8 1.108 = 7791.34
179 15. An object of 5 kg dropped from 16 m height. Calculate: i. Potential energy and kinetic energy possessed by the object before dropped. ii. Potential energy and kinetic energy possessed by the object after it dropped and landed on the ground. Answer: i. Potential energy and kinetic energy possessed by the object before dropped. Potential Energy, = ℎ = (5)(9.81)(16) = 784.8 Kinetic Energy, = 1 2 2 = 1 2 (5)(0) = 0 ii. Potential energy and kinetic energy possessed by the object after it dropped and landed on the ground. Potential Energy, = ℎ = (5)(9.81)(0) = 0 2 = 2 + 2 2 = 0 + 2(9.81)(16) 2 = 313.92 Kinetic Energy, = 1 2 2 = 1 2 (5)(313.92) = 784.8
180 16. Determine the velocity of the 105 kg crate as shown in figure below when it reaches the bottom of the chute at B. The initial velocity of the crate is 6 m/s at A and the friction force is 35 N. Answer: = = [ sin 15° − 35] × 10 = [(105)(9.81) sin 15° − 35] × 10 = (266.6 − 35) × 10 = 2316 = 2 − 1 = 1 2 2 − 1 2 2 2316 = 1 2 (105) 2 − 1 2 (105)6 2 2316 = 525 2 − 1890 2 = 2316 + 1890 525 2 = 80.11 = √80.11 = 8.95 /
181 17. An object weighs 2 kg falling from 10 m height. Calculate: i. Gravitational potential energy and the kinetic energy possessed by the object before it falls. ii. Gravitational potential and the kinetic energy possessed by the object right after it has touched the ground. Answer: i. Gravitational potential energy and the kinetic energy possessed by the object before it falls. Potential Energy, = ℎ = (2)(9.81)(10) = 196.2 Kinetic Energy, = 1 2 2 = 1 2 (2)(0) = 0 ii. Gravitational potential and the kinetic energy possessed by the object right after it has touched the ground. Potential Energy, = ℎ = (2)(9.81)(0) = 0 2 = 2 + 2 2 = 0 + 2(9.81)(10) 2 = 196.2 Kinetic Energy, = 1 2 2 = 1 2 (2)(196.2) = 196.2
182 18. Based on figure below, determine the tension developed in the cable when the motor, M winds in the cable with a constant acceleration. Then, the 25 kg crate moves a distance, = 6 in 4 , starting from rest. The coefficient of kinetic friction between the crate and the plane is = 0.25. Answer: = 35° = 25 × 9.81 × 35° = 200.9 = = 0.25 × 200.9 = 50.23
183 = − − 35° = 25 − 50.23 − (25 × 9.81 × 35°) = 25 × 0.75 − 190.9 = 18.75 = 18.75 + 190.9 = 209.65 given starting from rest = 0, = 6, = 4, = 0.25, = 25 = + 1 2 2 = (0)(4) + 1 2 (4 2 ) 6 = 8 = 6 8 = 0.75 m/s2
184 19. The 150 kg box as shown -in Figure 4(c) is originally at rest on the smooth horizontal surface. If a towing force of 300N acting at an angle of 45 0 is applied to the box for 8 s, determine the final velocity using: i. Equation of motion ii. Principle of work and energy Answer: i. Equation of motion + → ∑ = 300 45° = 150 = 300 45° 150 = 1.4142 m/s2 = + = 0 + (1.4142)(8) = 11.314 /
185 ii. Principle of work and energy = ( + ) 2 = (0 + )(8) 2 = 4 Principle of work and energy equation: = 2 − 1 = 1 2 2 − 1 2 2 = 1 2 (150) 2 − 1 2 (150)(0) (300 45°)(4) = 75 2 = 300 45° × 4 75 = 11.314 / 20. An 80 kg block as shown in figure below rests on horizontal plane. Determine the magnitude of the force P required ti give the block an acceleration of 2.5 m/s2 to the right. The coefficient of kinetic friction between the block and the plane is = 0.25.
186 Answer: Free Body Diagram (FBD) + ↑ ∑ =0 − sin 30° − = 0 = 30° + (80)(9.81) = 0.5 + 784.8 ----------------------------------- equation 1 + → ∑ = 30° − 0.25 = (80)(2.5) cos 30° − 0.25 = 200 ------------------------- equation 2 Substitute = 0.5 + 784.8 into equation 2 cos 30° − 0.25(0.5 + 784.8) = 200 0.866 − 0.125 − 196.2 = 200 0.741 = 200 + 196.2 0.741 = 396.2 = 396.2 0.741 = 534.68
187 21. The 100 kg crate shown in figure below is originally at rest on a smooth horizontal surface. If a force of 200 N is applied to the crate for 6 s, calculate the final velocity using: i. Equation of motion. ii. Principle of work and energy. Answer: i. Equation of motion. + → ∑ = 200 30° = 100 = 200 30° 100 = 1.732 m/s2 = + = 0 + (1.732)(6) = 10.392 /
188 iii. Principle of work and energy = ( + ) 2 = (0 + )(6) 2 = 3 Principle of work and energy equation: = 2 − 1 = 1 2 2 − 1 2 2 = 1 2 (100) 2 − 1 2 (100)(0) (200 30°)(3) = 50 2 = 200 30° × 3 50 = 10.392 / 24. A particle of 2 kg mass is being pulled across a smooth horizontal surface by a horizontal force. The force does 24 Joule of work in increasing the particle’s velocity from 5 m/s to v m/s. Calculate the value of v and the position of particle after 15 s. Answer: Work Done, = 2 − 1 = 1 2 2 − 1 2 2 24 = 1 2 (2) 2 − 1 2 (2)(5 2 ) 24 = 2 − 25 2 = 25 + 24 2 = 49 = √49 = 7 /
189 Find the acceleration: = + 7 = 5 + 15 = 7−5 15 = 0.133 m/s2 position of particle after 15 s = 1 2 ( + ) = 1 2 (7 + 5)15 = 90 25. A 50 kg crate rests on a horizontal plane for which the coefficient of kinetic () is 0.30. If the crate is subjected to a 400 N towing force as shown in figure below, determine the velocity of the crate in 5 s starting from rest. Answer: Free Body Diagram (FBD)