Phys 1401: General Physics I

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1 1. (0 Points) What course is this? a. PHYS 1401 b. PHYS 1402 c. PHYS 2425 d. PHYS (0 Points) Which exam is this? a. Exam 1 b. Exam 2 c. Final Exam 3. (0 Points) What version of the exam is this? a. A b. B c. C d. D 4. Which of these is equivalent to the SI unit of force, the newton? a. kg m/s b. kg m²/s c. kg m²/s² d. kg² m/s² e. kg m/s² F = ma, with m in kg, and a in m/s². 5. Which kinematics quantity is most closely tied to the concept of force? a. Position b. Displacement c. Velocity d. Speed e. Acceleration 6. Which of the following statements is correct? a. If an object currently has no forces acting on it, it cannot possibly be moving. b. If an object currently does have forces acting on it, it cannot possibly be stationary. c. Both (a) and (b) are true. d. Neither (a) nor (b) is true. With no forces, the object could be moving with constant velocity. With forces, they could cancel so that the net force is zero. 7. A bug hits the windshield of a fast-moving car. Which object experiences more force due to the impact? a. The bug. b. The car. c. They experience the same force. Newton s Third Law describes the forces as being equal. 8. Two books are stacked on a desk. Which is the proper description of the normal force acting on the bottom book and caused by the top book? a. Downward and equal to the weight of the top book. b. Upward and equal to the weight of the top book. c. Downward and equal to the combined weight of both books. d. Upward and equal to the combined weight of both books. e. Downward and equal to the weight of the bottom book. The bottom book is pressed downward by the Normal Force from top book. This is equal to the top book s weight when the books are at rest on a desk. 9. If a 5.0 kg mass is hanging at rest from a string, what is the tension in the string? a. 0.0 N b N c. 2.0 N d. 5.0 N e. 49 N F T F g = 0 F T = F g = mg = (5.0 kg)(9.8 N/kg) = 49 N

2 10. If a 5.0 kg mass is hanging from a string, moving upward at a constant 3 m/s, what is the tension in the string? a. 0 N b. 15 N c. 35 N d. 50 N e. 65 N With a constant velocity, a = 0 F T F g = 0 F T = F g = mg = (5.0 kg)(9.8 N/kg) = 49 N 11. An object is sitting on the flat floor of an elevator. The normal force exerted on the object by the elevator is affected by: a. The vertical velocity component. b. The horizontal velocity component. c. The vertical acceleration component. d. The horizontal acceleration component. e. Both (a) and (c). In the y-direction: F N F g = ma Only the vertical acceleration affects the normal force. 12. When an object is on an inclined ramp (0 < θ < 90 ), the magnitude of the normal force is a. 0 b. mg c. mg sin(θ) d. mg cos(θ) e. mg tan(θ) In the y direction: F N mg cos θ = A 3 kg block is sitting at rest on a ramp at an angle of 10. The coefficient of static friction between the block and the ramp is 0.5. What is the magnitude of the friction force acting on the block? a. 2.6 N b. 5.1 N c N d N e N In the x direction: F f mg sin θ = 0. F f = (3 kg)(9.8 N/kg) sin 10 = 5.1 N Using F f = μf N would tell us the maximum possible F f, but not the actual F f.

3 14. Three 2 kg masses are attached in a horizontal train using small strings. If a 12 N force is pulling the front mass forward, what is the tension force on the back mass? a. 12 N b. 8 N c. 6 N d. 4 N e. 3 N For the system: (12 N) = (6 kg)a, so a = 2 m/s 2. For the back mass: F T = (2 kg)(2 m/s 2 ) = 4 N. 15. When a car accelerates from rest to 30 m/s, what force propels the car forward? a. The friction force of the car on the road. b. The friction force of the road on the car. c. The normal force of the engine pushing on the car. d. The normal force of the engine pushing on the road. e. Gravity. 16. A block is on a frictionless ramp that is 1.5 m long. Starting from rest, the block is allowed to slide to the bottom of the ramp. What angle of the ramp will allow the block to reach the bottom in 1.1 s? a. 20 b. 15 c. 10 d. 5 e. 0 The average speed will be v avg = 1.5/1.1 = 1.36 m/s. The max speed will be 2v avg = 2.72 m/s. The acceleration is a = 2.72/1.1 = 2.48 m/s 2. For a frictionless ramp, mg sin θ = ma, so sin θ = 2.48/9.8 = and θ = A bicycle traveling at 5 m/s needs to stop in a distance of 3 m. What coefficient of friction is required? a. 8.3 b. 4.2 c d e To get the acceleration: v f 2 = v i 2 + 2aΔx, which leads to 5 2 = 2a(3), so a = 4.17 m/s 2. Horizontal motion: F f = ma = (4.17)m No vertical motion and assume flat road, so F N = F g = mg = (9.8)m. Coefficient of friction: μ = F f /F N = 4.17/9.8 = 0.43

4 18. A half-atwood machine is built out of a 10 kg mass on a frictionless horizontal surface. It is attached to a horizontal string that passes over a pulley, and the other end of the string hangs vertically with a 12 kg mass at the bottom. If the system starts from rest, what is the magnitude of its acceleration 0.1 s later? a. 18 m/s² b. 9.8 m/s² c. 8.2 m/s² d. 5.4 m/s² e. 4.5 m/s² For top block, x-direction: F T = m 1 a For bottom block, negative y-direction: mg F T = m 2 a Add to get system equation: m 2 g = (m 1 + m 2 )a Accel is a = (12)(9.8)/( ) = 5.35 m/s An Atwood s Machine is built from m 1 = 1 kg and m 2 = 4 kg. The massless string is passed over a frictionless, massless pulley. When the system is released from rest, what is the tension in the string? a. Less than 9.8 N b. 9.8 N c. Between 9.8 N and 39.2 N d N e. More than 39.2 N In an Atwood s machine: The light mass accelerates up, so F T must be more than its weight. F T > 9.8 N. The heavy mass accelerates down, so F T must be less than its weight. F T < 39.2 N. 20. Which is a correct equation of motion for a the diagram shown? Assume the motion is in the indicated direction, and the system was released from rest. There is a friction coefficient of μ k between m 2 and the ramp. a. +m 2 g cos(θ) μ k m 2 g sin(θ) + m 1 g = (m 1 + m 2 )a b. +m 2 g sin(θ) + μ k m 2 g cos(θ) m 1 g = (m 1 + m 2 )a c. m 2 g cos(θ) + μ k m 2 g sin(θ) + m 1 g = (m 1 + m 2 )a d. m 2 g sin(θ) μ k m 2 g cos(θ) m 1 g = (m 1 + m 2 )a e. +m 2 g sin(θ) μ k m 2 g cos(θ) m 1 g = (m 1 + m 2 )a +m 2 g sin θ is the only forward force. m 2 g cos θ is the normal force on block 2, so μ k m 2 g cos θ is the backward friction force m 1 g is a backward force 21. The density of ice is 917 kg/m³. The density of seawater is about 1030 kg/m³. What percentage of the volume of a floating iceberg is sticking up above the surface of the water? a. 89% b. 50% c. 12% d. 11% e. 0% F g = ρ ice V ice g = F B = ρ water V under g The fraction under water is V under = ρ ice = 917 = 0.89 = 89%. V ice ρ water 1030 The fraction above water is (1 0.89) = 0.11 = 11% = 100% 89%.

5 22. A magnesium block has a weight of 34.8 N. When the block is suspended under water by a string, the tension in the string is 14.8 N. What is the density of magnesium? a kg/m³ b kg/m³ c kg/m³ d kg/m³ e kg/m³ F g = 34.8 N = ρ Mg V Mg g F B = ( ) = 20 N = ρ water V Mg g So the specific gravity is ρ Mg ρ water = = Since water has a density of 1000 kg/m³, magnesium has ρ Mg = 1740 kg/m A heavy ball is held up by two strings. One is horizontal while the other is tilted at an angle of 60 from the horizontal. The tension in the diagonal string (T 1 in the diagram) is 50 N. What is the weight of the ball? a. 25 N b. 29 N c. 43 N d. 50 N e. 87 N T 1 is 60 from horizontal, so cos(θ) goes with the horizontal component. Horizontal: T 2 (50) cos 60 = 0 so T 2 = 25 N (not asked for). Vertical: (50) sin(60 ) T 3 = 0, so T 3 = 43.3 N. Since the system is at rest, T 3 is the weight of the ball. 24. An ideal meter stick is balanced on a pencil which is at the 30 cm mark. A 150 g mass sits at the 20 cm mark. What is the mass of the meter stick? a. 50 g b. 75 g c. 90 g d. 150 g e. 300 g The pivot point is the pencil. The meter stick c.g. is at 50 cm, which is 20 cm away. The 150 g mass is at 20 cm, which is 10 cm away. τ stick + τ 150 = 0, so τ stick = τ 150 m stick g(20 cm) =(150 g)g(10 cm) m stick = 75 g Don t confuse g for grams with g for gravity.

6 25. The free body diagram below represents a 1500 kg car sitting on a 3000 kg bridge supported at its far ends. The car's position is three quarters of the length L from the left end of the bridge. Identify the one error in the torque equation: F 1 L F g (L/2) F car (3L/4) + F 2 L = 0 (Hint: figure out where the pivot point should be so that there is only one error in the equation.) a. F 1 L should be negative. b. F 1 L should be 0. c. F 2 L should be 0. d. F car (3L/4) should be F car (L/4) e. There is no error. The pivot point can t be L from F 1 and F 2 simultaneously, so one of those must be wrong. The pivot point is 3/4 of the way from the car, so the pivot point is at F 1. Therefore, F 1 cannot exert a torque, and F 1 L should be zero. 26. A student has drawn a free-body diagram for a ladder of mass m leaning against a frictionless wall. The diagram is shown below. What is the error in the diagram? a. f s is in the wrong direction. b. F wall should be directed upwards, not down. c. F wall should be perpendicular to the wall, not parallel to it. d. n should be down into the floor for f s to have the given direction. e. F wall should be diagonal or split into components. Since the wall is frictionless, it exerts a normal force and not a friction force. Therefore, F wall should be perpendicular to the wall. 27. A diagonal bar is pinned at the origin and extends up to (6 m)i + (8 m)j. Its weight is 10 N. How much torque about the pin does the gravitational force of Earth cause? (As usual, j is the vectical direction.) a. 10 N m b. 30 N m c. 40 N m d. 60 N m e. 100 N m For a vertical force, the perpendicular distance is horizontal. Gravity is exerted at the midpoint, which is horizontally 3 m from the origin. So the torque due to gravity is (10 N)(3 m) = 30 N m.

7 28. A uniform beam having a mass of 60 kg and a length of 2.8 m is held in place at its lower end by a pin. Its upper end leans against a vertical frictionless wall as shown in the figure. What is the magnitude of the force the pin exerts on the beam? a kn b kn c kn d kn e kn Call the components of the pin force F Px and F Py. In the vertical direction: F Py mg = 0, so F Py = (60)(9.8) = 588 N. In the horizontal direction: F Px F N = 0, so F Px = F N. Looking at torques, gravity is exerted at the midpoint and is 50 from the bar. τ g = F g r g sin(θ g ) = (60 9.8)(1.4) sin(50 ) = 631 N m. The wall force is exerted at the end and is 40 from the bar. τ τ N = F N r N sin(θ N ), so F N = N = 631 = 350 N r N sin(θ N ) 2.8 sin(40 ) Combined, the force of the pin is: F P = = 684 N