Name Date Pd Forces and Laws of Motion Worksheet 4.6 –Applications of Newton’s Laws For each of the problems below, you must begin your solution with a free body diagram 1. A 20.0 kg crate hangs at the end of a long rope. Find the size and direction of its acceleration when the tension in the rope is a) 250 N b) 150 N FT=250N a FT=150N a Fg=mg=196N Fy ma y FT Fg ma 250 196 20a a 2.7 m / s 2 up Fg=mg=196N Fy ma y FT Fg ma 150 196 20a a 2.3m / s 2 down 2. In a modified tug of war game, two people pull in opposite directions, not on a rope, but on a 25.0 kg sled resting on an icy road. If the participants exert forces of 90.0 and 92.0 N, how much will the sled accelerate? Fx max 92 90 25a a 0.08m / s 2 in same direction as 92 N force 3. Compute the size of the initial upward acceleration of a rocket of mass 1.30 x 104 kg if the initial upward thrust of its engine is 2.60 x 105 N. Do not neglect the weight of the rocket. Construct the force diagram Fthrust =260,000 N Fy ma y FThrust Fg ma y 260,000 127,400 13,000a a 10.2m / s 2 up a Fg=mg=127,400N 4. A 60 kg woman stands in an elevator that is accelerating downward at 3.10 m/s2. a) Construct the force diagram for the woman b) What is her weight? FN Fg mg 588N c) What is her apparent weight? (measured by the normal force acting on her) a=3.10 F ma y FN Fg FN FN FN y ma y Fg ma 588 60(3.1) 402N Fg=mg=588N d) The elevator now accelerates up at 3.10 m/s2. Construct the force diagram and determine the woman’s apparent weight. Fy FN Fg FN FN FN ma y ma y Fg ma 588 60(3.1) 774N FN a=3.10 Fg=mg=588N 5. The maximum force that a grocery bag can withstand without ripping is 250 N. Suppose that bag is filled with 20.0 kg of groceries and lifted with an acceleration of 5.0 m/s2. Do the groceries stay in the bag? Flift Fy Flift Fg Flift Flift Flift ma y ma y Fg ma 196 20(5) 296 N a The bag rips Fg=mg=196N 6. Suppose that a 1000 kg car is traveling at 25 m/s (55 mph). Its brakes can apply a force of 5000N. What is the minimum stopping distance required for the car to stop? v0x=25 m/s vfx=0 Fbrake=5000N Fx max Fbrake ma y 5000 1000a a 5m / s 2 Dx=? v 2fx v02x 2aDx 0 252 2(5)(Dx) Dx 62.5m 7. Below is a picture of an Atwood’s machine: two masses attached to a frictionless, massless pulley. The mass of block A is 5.0 kg and the mass of block B is 2.0 kg. a) What is the acceleration of the system when the blocks are released? See the lecture powerpoint for how to solve this problem by applying Newton’s 2nd Law to each block and using the 2 equations to solve for the two unknowns (FT and a). FT FT a A B FgA = mg = 49N a Here I will show how to solve the problem as a system. If we consider block A and B as a system of mass mA + mB, then the force of gravity pulling block A down, FgA, is opposed only by the tension, FT which is the force of gravity pulling on B, FgB, since FT’s are internal forces that cancel to 0. Newton’s 2nd Law for the SYSTEM: FgB = mg = 19.6N FgB = mg = 19.6N B Fsys y msys a y FgA FgB (m A mB )a y 49 19.6 (5 2)a y a y 4.2m / s 2 a FT FT A FgA = mg = 49N Internal forces External forces b) How long will it take for block A to fall 2.0 m? v0 y 0 v fy a y 4.2m / s 2 Dy 2m t Dy v0 y t 12 a y t 2 2 12 (4.2)t 2 t 0.98s 8. A shuffleboard disk is accelerated to a speed of 5.8 m/s and released. If the coefficient of kinetic friction between the disk and the concrete court is 0.31, how far does the disk go before it comes to a stop? The courts are 15.8 m long. FN fK Vfx=0 V0x=5.8m/s Dx=? Fg Need to find acceleration. Only force acting in the direction of motion(x) is friction. Fx max f k max k FN max (0.31)( Fg ) max (0.31)(mg ) max a x (0.31)(9.8) a x 3.04m / s 2 v 2fx v02x 2aDx 0 5.82 2(3.04)(Dx) Dx 5.53m 9. Stacie, who has a mass of 45 kg, starts down a slide that is inclined at an angle of 45° with the horizontal. If the coefficient of kinetic friction between Stacie’s shorts and the slide is 0.25, what is her acceleration? x-axis is parallel to the incline and y-axis is perpendicular F N Fx max f k Fgx max k FN mg sin 45 max k Fgy mg sin 45 max k mg cos 45 mg sin 45 max (0.25)(9.8)(0.707) (9.8)(0.707) a x a x 5.2m / s 2 down the incline fK Fgy 45o 45o Fgx Fg=mg=441N 10. A 63-kg water skier is pulled up a 14.0° incline by a rope parallel to the incline with a tension of 512 N. The coefficient of kinetic friction is 0.27. What are the magnitude and direction of the skier’s acceleration? FN x-axis is parallel to the incline and y-axis is perpendicular Fx max f k Fgx FT max k FN Fg sin 14 512 max k Fgy Fg sin 14 512 max k Fg cos 14 Fg sin 14 512 max (0.27)(617.4)(0.970) (617.4)(0.242) 512 (63)a x a x 3.19m / s 2 14o 14o Fgy up the incline Fg=617.4N Fgx 11. A 15 kg block is on a ramp that is inclined at 20° above the horizontal. It is connected by a string to a 19 kg mass that hangs over the top edge of the ramp. The coefficient of kinetic friction between the incline and the 15 kg block is 0.22. What is the magnitude of the acceleration of the 19 kg block? (See the figure.) FN a(+) FT FT a(+) fk Fgy Fg1=mg Fgx Fg2=mg =186.2N 15 kg Block on incline: Fx f K Fgx FT k FN mg sin 20 FT (0.22) Fgy (15)(9.8) sin 20 FT (0.22)(mg cos 20) 50.28 FT 30.39 50.28 FT FT 19 kg hanging block: max max 15a 15a 15a 15a 15a 80.67 Fy ma y FT Fg ma FT 186.2 19a FT 19a 186.2 15a 80.67 19a 186.2 34a 105.53 a 3.1m / s 2 down You can also solve the problem by treating the 2 blocks as one system since they are connected by one rope and accelerate the same. Then the forces of tension are internal forces that cancel out within the system. The force of gravity on the 19kg block is opposed by the forces parallel to the incline (fK and Fgx) For the system: Fy msys a y a(+) FN Fg 2 f K Fgx (15 19)a y 186.2 k FN mg sin 20 34a y FT FT a(+) fk 186.2 (0.22) Fgy (15)(9.8) sin 20 34a y Fgy 186.2 (0.22)(mg cos 20) 50.28 34a y Fg1=mg 186.2 30.39 50.28 34a y Fgx a y 3.1m / s 2 12. In the figure to the right, two boxes of masses m=2kg and 4m=8 kg are in contact with each other on a frictionless surface. A force of F=50 N is applied to m. a) What is the acceleration of the more massive box? Both boxes accelerate the same so treat both boxes as a system: Fg2=mg =186.2N Fx msys a x Fpush (5m)a 50 (2 8)a a 5m / s 2 b) What is the force causing the acceleration of the more massive box (the net force on 4m)? The force causing the acceleration of the larger box is the force applied from the smaller box, F4m by m. For the larger box: Fx max F4 m by m (8)(5) 40 N F4m,m F = 50N m 4m c) What is the force exerted on the smaller box by the larger box? It is equal and opposite to the force exerted on the larger box by the smaller box: 40 N For the smaller box: Fx max F Fm by 4 m 2a 50 Fm by 4 m (2)(5) Fm by 4 m 40 N F = 50N Fm,4m m 4m 4m d) What is the net force on the smaller box? m Fx F Fm by 4m 50 40 10 N F = 50N F4m,m // // Fm,4m \ e) Show all the action reaction pairs of forces for all the forces acting on the boxes. Action-reaction force pairs are shown in the same color. For every action there is an equal and opposite reaction. Remember action and reaction forces act on different objects and therefore do not cancel out FN (on 4m by table) FN (on m by table) F(on m by 4m) F(on 4m by m) FN (on table by m) Fg (on m by E) FN (on table by 4m) Fg (on 4m by E) Fg (on Earth by m) Fg (Earth by 4m) f) with friction (k = 0.2) Both boxes accelerate the same so treat both boxes as a system: Fx msys a x Fpush f k (5m)a F k FN a push 5m 50 0.2 Fg 50 0.2(10 g ) 10 10 2 a 3.04m / s FN System 4m m F = 50N fk Fg Net force on 4m box: Since its acceleration is 3.04m/s in +x direction: Fx Fnet 4max Fnet 8(3.04) 24.3N Force exerted on the smaller box by the larger box? Fx m1a x F12 F F12 f k max fk F12 F f k max F k FN 1 max 50 0.2(2)(9.8) 2(3.04) F12 40 N FN1 F = 50N 4m m Fg1 net force on the smaller box. Since its acceleration is 3.04m/s in +x direction: Fx Fnet max Fnet 2(3.04) 6.08 N 13. This arrangement is being pulled across the floor. The coefficient of kinetic friction between the larger block and the floor is (0.50). The coefficient of static friction between the two blocks is (0.75). (Note that the weight, not the mass, of each block is given) a) What is the maximum acceleration of the blocks before the top block begins to slide off the bottom block? (Hint: Draw the free body diagram of the top block and determine the acceleration when the coefficient of static friction is at its max value of 0.75 and the top block is at rest relative to the bottom block) F12 For the blocks to move together, the top block is held in place by static friction. Without friction between the blocks, the top block would slide off the back of the bottom block. Therefore, the force of static friction is fs1 30N opposite this motion, or pointing to the right. At right is shown the FBD of the top block. The force on the Fg1 top block by the bottom (F12) acts as a normal force. Max acceleration of the blocks before top slides off is 100N when fs1 is maximum. Net force on top block: Fx f s1 s F12 s m1 g ax m1a x m1a x m1a x m1a x s g 7.35m / s 2 F b) What is the maximum horizontal Force, F, that can be exerted on the lower block before the top block begins to slide off? (Hint: Draw the free body diagram of the bottom block and apply Newtons 2nd law to the bottom block) The max force is the force that results in an acceleration of 7.35m/s2, found in part a). FN Note that F21=F12=Fg1 = 30N. Net force on the bottom block: Fx m2 a x F f k 2 m2 a x F k FN m2 a x F k ( Fg 2 F21 ) m2 a x F m2 a x k ( Fg 2 F21 ) 10.2(7.35) 0.5(100 30) F 140 N 30N F 100N fk2 F21 Fg2 Fy 0 FN Fg 2 F21 0 FN Fg 2 F21 100 30 130 N (Normal force bears the weight of both boxes) 14. Two blocks of mass m1 = 2 kg and mass m2 = 7 kg are wedged up against one another and against a wall by a horizontal force F. Doing each section algebraically before putting in numbers: a) What is the coefficient of static friction between m1 and m2 (call this s1) and between m2 and the wall (call this s2) if the MINIMUM force Fmin required to keep the blocks from breaking loose and sliding under the influence of gravity if Fmin = 25 N? m1 m2 F To find the friction between the blocks and the friction between the block and wall, need to look at each block separately. The freebody diagrams for each block are shown below. In each case, the force balancing gravity and preventing the blocks from falling is static friction. fs1 F12 // m1 fs2 m2 F // wall wall m1 F FN // m2 Fg1 Fg2 F21 // wall Block 1 Block 2 Fy m1a y 0 Fy m2 a y 0 Fg1 f s1 0 Fg 2 f s 2 0 m1 g s1 FN 0 m2 g s 2 FN 0 s1 F12 m1 g s1 s 2 F m2 g m1 g m1 g F12 F m2 g F 2.74 s2 s2 s1 0.784 b) The force F is decreased to 20 N. The blocks break loose and begin to fall. If the coefficient of kinetic friction between m1 and m2 AND between m2 and the wall are k1 = 0.15 and k2 = 0.9 respectively, what are the accelerations of m1 and m2? Note that they will not be the same. fk1 F12 // m1 m2 fk2 wall F // wall m1 F FN m2 // F21 // Fg1 Fg2 Block 1 Block 2 Fy m1a1 y 0 Fy m2 a2 y 0 Fg1 f k1 m1a1 y Fg 2 f k 2 m2 a2 y m1 g k1 FN m1a1 y m2 g k 2 FN m2 a2 y a1 y m1 g k1 F12 m1 m1 g k1 F m1 a1 8.3m / s 2 (down) a2 y m2 g k 2 FN m1 m2 g k 2 F m2 a2 7.23m / s 2 (down) c) Why are the accelerations different? The coefficient and force of kinetic friction acting on each of the blocks is different and this causes the acceleration of each block to be different 15. A block is set on an adjustable ramp and the angle of incline is increased slowly until the block is observed to start sliding. (a) Solve for the coefficient of static friction, μs, in terms of the maximum angle of incline, θ, at which the block can remain at rest on the ramp. Fx max 0 FN fs Fgx f s 0 Fg sin q s FN 0 mg sin q s Fgy 0 Fg2x mg sin q s mg cos q 0 mg sin q sin q mg cos q cos q s tan q s q Fgy q Fg=mg (b) Considering the fact that μs is almost always less than or equal to one, at what amount of incline would there be too little friction to prevent virtually anything from sliding? At q = 90o, since tanq = 1

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