Kinematics and Dynamics

Mr. Klapholz Shaker Heights High School Kinematics and Dynamics Problem Solving

Problem 1 A person travels from x = 20 m to x = 11 m. The trip takes 18 s. What is the person’s displacement, speed and velocity?

Solution 1 (1 of 2) s = Dx s = xf – xi s = 11 m – 20 m s = -9 m

Solution 1 (2 of 2) s = -9 m speed = distance / time speed = 9 m / 18 s speed = 0.5 m s-1 v = s ÷ t v = (-9 m) ÷ (18 s) v = -0.5 m s-1 In physics, the minus sign is always telling us about the direction. The magnitude of v is speed.

Problem 2 A car is moving at 30 m s-1 and slows to a stop (“rest”) in 10 s. What is the average acceleration of the car?

Solution 2 a = Dv ÷ t a = (v2 – v1) ÷ t a = ( 0 m s-1 – 30 m s-1 ) ÷ (10 s) a = (-30 m s-1 ) ÷ (10 s) a = -3 m s-2 This means that the car is slowing at 3 m/s every second. Acceleration is the rate of change of velocity.

Problem 3 If you climb to a high place and drop a rock, how far does the rock fall in 2 seconds?

Solution 3 s = ut + (1/2)at2 s = (0 m s-1)(2 s) + (1/2)(-9.8 m s-2)(2 s)2 s = -19.6 m The rock travels a distance of 19.6 m

Problem 4 If you climb to a high place and throw a rock upwards at 5 m s-1, then where will the rock be after 2 s?

Solution 4 s = ut + (1/2)at2 s = (5 m s-1)(2 s) + (1/2)(-9.8 m s-2)(2 s)2 s = 10 m - 19.6 m The rock will be -9.6 m below your position.

Problem 5 If you climb to a high place and throw a rock downwards at 5 m s-1, then where will the rock be after 2 s?

Solution 5 s = ut + (1/2)at2 s = (-5 m s-1)(2 s) + (1/2)(-9.8 m s-2)(2 s)2 s = -10 m - 19.6 m s = - 29.6 m The rock will be 29.6 m below your position.

Problem 6 If you throw a ball upwards at 30 m s-1, then when will the ball return to your hand? Take the acceleration of gravity to be 10 m s-2.

Solution 6 s = ut + (1/2)at2 0 = (30 m s-1)t + (1/2)(-10 m s-2)t2 0 = 30t + -5t2 5t2 = 30t t = 0 solves the equation. 5t = 30 t = 6 s

Problem 7 If you throw a ball upwards at 30 m s-1, then what is the greatest height that the ball will reach? Take the acceleration of gravity to be 10 m s-2.

Solution 7 Method One At the top, v = 0. v2 = u2 + 2as (0)2 = (30)2 + 2(-10)s 0 = 900 – (20)s 20s = 900 s = 900 / 20 s = 45 m

Solution 7 Method Two We know that the ball will be back at the hand in 6s. So, at t = 3 s, the ball is at its maximum height. s = (30)(3) + (1/2)(10)32 s = 90 - 45 s = 45 m

Problem 8 For the graph, find the velocity at 6 seconds. The original velocity is 1 m s-1. 10 a / m s-2 2 0 6 0 t / s

Solution 8 Background: The area under a velocity graph is the change in position. The area under an acceleration graph is the change in velocity. So, the final velocity = initial velocity + area. The area has two parts: the triangular part and the rectangular part:

Solution 8 10 a / m s-2 Area 1 2 Area 1 0 6 0 t / s

Solution 8 The area of the triangle is (1/2)Base•Height Area of Triangle = (1/2)(6 s)(8 m s-2 ) = 24 m/s Area of Rectangle= Base•Height Area of Rectangle= (6 s)(2 m s-2 ) = 12 m s-1 Total Area + 36 m s-1 New Velocity = 1 m/s + 36 m/s = 37 m/s

Problem 9 What is the weight of a 3 kg bag of potatoes?

Solution 9 Weight is the force that the earth puts on an object. If the object is at the surface of the earth, then the weight is the product of the mass and the acceleration of gravity. W = mg Weight = (3 kg)(9.8 m s-2) W = 29.4 Newtons W = 29.4 N

Problem 10 A 2 kg bag of sugar is being pulled across a tabletop with a fore of 24 N. You might think that the acceleration would be 12 m s-2, but it isn’t. The acceleration is 8 m s-2. What is the force of friction?

Solution 10 Draw the situation. SF = ma 24 N - friction = (2 kg)(8 m s-2) 24 N – friction = 16 N 24 N – 16 N = friction friction = 8 N

Problem 11 There is a boat that travels at 2.0 m s-1, but that only applies if the water is not moving. We say “It’s speed is 2.0 m/s in still water.” This boat is in a river that flows south at 5.0 m/s. The boat is on the East bank of the river. The boat uses its compass to aim due West. The velocity of the boat, relative to the water, is 2 m/s West. The velocity of the water, relative to the earth, is 5 m/s South. What is the velocity of the boat relative to the earth?

Solution 11 (1 of 2) Draw the situation. Do you see that the speed of the boat must be more than 5 m/s, but less than 7 m/s? Velocity is speed and direction, so we’ll need both. We can find the speed using the Pythagorean theroem: v2 = 5.02 + 2.02 v2 = 25 + 4 v2 = 29 v = 5.4 m s-1 (Notice the significant figures.)

Solution 11 (2 of 2) We still need the direction of the boat’s velocity with respect to the earth. q = InvTan ( Opposite / Adjacent ) q = ArcTan ( 2.0 m/s / 5.0 m/s ) q = Tan-1 ( 0.40 ) q = 22˚ … v = 5.4 m/s, 22˚ West of South Or v = 5.4 m/s, 68˚ South of West

Problem 12 A train starts from rest and accelerates uniformly to a speed of 45 m s-1 in 3 minutes (180 s). The train travels at this speed for 4 min. (240 s), and then it slows, with a constant acceleration, to rest in 2 minutes (120 s). Find the accelerations. Find the distance traveled.

Solution 12 (1 of 3) Draw a graph of velocity against time. Optional: draw a graph of position vs. time. Acceleration is the gradient of the velocity graph. The first segment has a slope of: a = Dv / t a = (45 m/s - 0 m/s) / 180 s a = (45 m s-1) ÷ (180 s) a = 0.25 m s-2 (sig fig)

Solution 12 (2 of 3) The second segment has no slope, so no acceleration. The third segment has a slope of: a = Dv / t a = (0 - 45) / 120 s a = ( -45 m s-1) ÷ (120 s) a = -0.38 ms-2 Why is this acceleration greater?

Solution 12 (3 of 3) Displacement is the area under the velocity curve. Area = Area of Triangle + Rectangle + Triangle Area One = (1/2)Base×Height = (1/2)(180s)(45m/s) = 4050 m Area Two = Base×Height = (240s)(45m/s) = 10800 m Area Three = (1/2)Base×Height = (1/2)(120s)(45m/s) = 2700 m Total Area = 4050 + 10800 + 2700 m = 17550 m Two Significant Digits: Displacement = 18000 m

Problem 13 An object starts at rest and speeds up. The rate at which it speeds up is changing. The acceleration starts at zero, and increases uniformly to 30 m s-2 at 10 s. How fast is it moving at 10 s?

Solution 13 (1 of 2) The acceleration is not uniform. Sketch the graph of acceleration vs. time. The area under the graph is the change in velocity. Area of a Triangle = (1/2)Base×Height Change in Velocity = (1/2)(10 s) × (30 m s-2) Dv = 150 m s-1 Since the object started from rest, the final speed is 150 m/s. Extension…

Solution 13 (2 of 2) Extension What if the problem was the same, except the original velocity was not 0, but instead +10 m s-1? The change in velocity would still be 150 m/s, but the new velocity would now be… 160 m s-1 We just used the equation: New Value = Old Value + Change in Value

Problem 14 A boat is crossing the Amazon River. The river is 110 m wide, and flows Northward. The river flows at 3.0 m s-1. The boat, if it had been in ‘still’ water, would travel at 4.0 m s-1. The boat is traveling toward the East. How far ‘downriver’ does the boat travel as it crosses? What is the velocity of the boat, relative to the earth?

Solution 14 (1 of 3) Draw the situation.

Solution 14 (2 of 3) How far does the boat travel downriver, as it crosses? Well, it travels downriver at 3.0 m/s. Distance = Speed•Time = ( 3.0 m s-1 ) • ( ? ) How long does it take to cross? Time = Distance ÷ Speed Time = (110 m) / (4.0 m/s) = 27.5 s Distance Downriver = 3 m/s • 27.5 s = 82.5 m D = 83 m

Solution 14 (3 of 3) Use the drawing. To get the speed, use the Pythagorean theorem: v2 = 3.02 + 4.02 v = 5.0 m s-1 To get the direction: Tan q = Opposite ÷ Adjacent q = tan-1( 4.0 ms-1 / 3.0 ms-1 ) = 53˚ East of North Or, q = Invtan(3.0ms-1/4.0 ms-1) = 37˚ North of East So, v = 5.0 m s-1, 37˚ North of East

Problem 15 Draw “Free-Body Diagrams” of: an airplane flying at constant velocity, And a car slowing down for a traffic light.

Problem 16 Two forces act on a 10 kg object. Find the acceleration of the object. One of the forces is in the negative x direction, and has a magnitude of 6.0 N. The other force is at an angle of 30˚ to the positive x axis, and has a magnitude of 4.0 N.

Solution 16 (1 of 3) Draw the situation. What are the components of the forces? The first force has no y component, and an x component of – 6.0 N. F1x = -6.0 N F1y = 0 N The second force is more complicated. The x component is adjacent to the 30˚ angle, so F2x = (4.0 N)cos30˚ F2y = (4.0 N)sin30˚ F2x = 3.5 N F2y = 2.0 N

Solution 16 (2 of 3) F1x = -6.0 N F1y = 0 N F2x = 3.5 N F2y = 2.0 N The total force has an x comp. of -6.0 + 3.5 = -2.5 N The total force has an y component of 0 + 2.0 = 2.0N F2 = Fx2 + Fy2 = (2.5)2 + (2.0)2 = 10.25 F = 3.2 N a = SF / m a = 3.2 N / 10 kg = 0.32 m s-2

Solution 16 (3 of 3) Direction. The total force has an x component of 2.5 N The total force has an y component of 2.0 N The angle between the total force and the –x axis is q = InvTan ( Opp / Adj ) q = ArcTan ( 2.0 / 2.5 ) = 39˚ So the acceleration is a = 0.32 m s-2 , 39˚

Problem 17 A 1.0 kg block of frictionless ice is on a smooth tabletop. A string is attached to the side of the ice, and the other end of the string is attached to a 0.50 kg block of tofu. The tofu is hanging off of the edge of the table. When the system is released, what is the acceleration of the tofu?

Solution 17 Draw the situation. The force that drives the system is the weight of the tofu: Wtofu = mtofug The inertia of the system is: mtofu+Mice SF = ma mtofug = { mtofu + Mice } a a = mtofug ÷ {mtofu+Mice} a = (0.50)(9.8) ÷ {0.50+1.0} a = (4.9) ÷ {1.50} = 3.3 m s-2 a = 3.3 m s-2

Problem 18 A person (70.0 kg) is traveling in a car at highway speed (30.0 m s-1). There is a problem 180.0 m ahead! How much force will it take to stop the person?

Solution 18 SF = ma F = (70.0 kg)•a But what is the acceleration? Which equation from kinematics is best for this job? v2 = u2 + 2as 02 = 30.02 + 2a(180.0) a = -302 ÷ 2(180) = -2.50 m s-2 F = (70.0)(-2.50) F = -175 N

Tonight’s HW: Go through Mechanics section in your textbook and scrutinize the “Example Questions” and solutions. Bring in your questions to tomorrow’s class.

Kinematics and Dynamics