CHAPTER 4NEWTON’S LAWS OF MOTION INTERNET ASSIGNED PROBLEMS 1 – 30 Isaac Newton (1642 – 1727) NEWTON’S LAWS: SECOND & THIRD 1. What force is needed to accelerate a child on a bike (total mass = 60.0 kg) at 1.25 m/s2? Hint: Use Newton’s second law to calculate force. 2. A net force of 265 N accelerates a bike and rider at 2.30 m/s2. What is the mass of the bike and rider together? Hint: Use Newton’s second law to calculate mass. 3. How much tension must a rope withstand if it is used to accelerate a 960-kg car horizontally along a frictionless surface at 1.20 m/s2? Hint: Force and tension mean the same thing. Use Newton’s second law to calculate tension. 4. What average force is required to stop an 1100-kg car in 8.0 s if the car is traveling at 95 km/h? Hint: Convert km/h to m/s by dividing it by 3.6. Use Newton’s second law to find force. 5. What average force is needed to accelerate a 7.00-gram projectile from rest to 125 m/s over a distance of 0.800 m along the barrel of a rifle? Hint: The average force on the pellet is its mass times its average acceleration. Convert the grams to kilograms. 6. A fisherman yanks a fish vertically out of the water with an acceleration of 2.5 m/s2 using a fishing line that has a breaking strength of 22 N. The fisherman loses the fish as G the line snaps. FT What was the minimum mass of the fish? Hint: We assume that the fish line is pulling vertically on the fish. G mg 7. A 0.140-kg baseball traveling 35.0 m/s strikes the catcher’s mitt, which, in bringing the ball to rest, recoils backwards 11.0 cm. what was the average force applied by the ball on the glove? Hint: By Newton’s 3rd law, the force exerted by the ball on the glove is equal and opposite to the force exerted by the glove on the ball. So calculate the average force on the ball, and then take the opposite of that result to find the average force on the glove. 8. A 12.0-kg bucket is lowered vertically by a rope in which there is 163 N of tension at a given instant. What is the acceleration of the bucket? Hint: Solve using Newton’s second law. 9. An elevator (mass 4850 kg) has a maximum acceleration of 0.68g. Hint: To find the maximum tension, assume that the acceleration is up. What is the maximum force the motor should exert on the cable? 10. A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.75 of the person’s weight. Calculate the acceleration of the elevator. G mg Hint: There will be two forces on the person – their weight, and the normal force of the scales pushing up on the person. G FN 11. The cable supporting a 2125-kg elevator has a maximum strength Hint: Choose UP to be the positive direction. Use Newton’s second law to solve. of 21,750 N. What maximum upward acceleration can it give the elevator without breaking the cable? G FT G mg 12. On planet Alpha an astronaut picks up a rock which has a mass of Hint: Apply Newton’s Second Law to find 5.00-kg and its weight on the planet is 40.0 N. If the astronaut the acceleration of the rock. exerts an upward force of 46.2 N on the rock, what is its acceleration? G F 13. If a 10.3-kg cart at rest is pushed with a force of 12.1 N. How far does it travel in 2.50 s? Hint: Use Newton’s Second Law to find the acceleration of the cart, then use the acceleration to find the distance. 14. A 92-kg water skier floating on a lake is pulled from rest to a speed of 12 m/s in a distance of 25 m. What is the net force exerted on the skier? Hint: Determine the acceleration of the skier, then use Newton’s Second Law to find find the force exerted on the skier. NEWTON’S LAWS: VECTORS FRICTION & INCLINED PLANES 15. A tractor tows a 3900-kg trailer up a 16o inclined with a speed of 3.0 m/s. y G F x θ Hint: Because the trailer is moving at constant speed, the net force on the trailer must be zero, and the force exerted by the tractor on the trailer must equal the component of the trailer’s weight that is pointing down the incline and parallel to it. G mg mg sin θ What force does the tractor exert on the trailer? 16. Two crewman pull a raft past a dock, as shown. One crewman pulls with a force of 30.0 N at an angle of 34o relative to the forward direction of the raft. The second crewman, pulls with force of 40.0 N at an angle of 45o. Hint: Use the vector sum of the forces to find the resultant force. What is the sum of the forces exerted by both crewmen? 17. A person pushes a 14.0-kg lawn mower at constant speed with a force of 88.0 N directed along the handle, which is at an angle of 45o to the horizontal as shown. Hint: The forces must sum to 0 since the mower is not accelerating. Calculate the horizontal friction force. 18. Based upon the information in problem 17 calculate the normal force exerted vertically upward on the mower by the ground. Hint: The forces must sum to 0 since the mower is not accelerating in the vertical direction. 19. What force must the person in problem 17 exert on the lawn mower to accelerate it from rest to 1.5 m/s in 2.5 seconds, assuming the same friction force? Hint: This requires two equations to solve. 20. If the coefficient of kinetic friction between a 35-kg crate and the floor is 0.30, what horizontal force is required to move the crate at a steady speed across the floor? Hint: The crate does not accelerate vertically, and the crate does not accelerate horizontally. G FN G FP G Ffr G mg 21. A force of 48.0 N is required to start a 5.0-kg box moving across Hint: To start the box moving, the pulling force must just overcome the force a horizontal concrete floor. What is the coefficient of static friction between the box and the floor? of static friction. 22. If you are standing on a train that is accelerating at 0.20g, what minimum coefficient of static friction must exist between your feet and the floor if you are not to slide? Hint: The maximum static frictional force is μ s FN and that must be greater than or equal to the force needed to accelerate you. G Ffr G mg G FN 23. What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the ground is 0.80? G FN Hint: The car does not accelerate vertically. The static frictional force is the accelerating force, and so Ffr = ma . G Ffr G mg 24. A baseball player slides into third base with an initial speed of 4.0 m/s. If the coefficient of kinetic friction between the player and the ground is 0.46, how far does the player slide before coming to rest? Hint: Find the acceleration of the player and insert into the distance equation. 25. A child goes down a playground slide with an acceleration of. 1.16 m/s2. Find the coefficient of kinetic friction between the child and the slide if the slide is inclined at an angle of 31.0o below the horizontal. Hint: Choose the x-axis along the direction of motion. Write Newton’s Second Law in the y direction to find the normal force, and then write Newton’s Second Law in the x direction and solve for µk. 26. You floor your Honda Civic and accelerate at 12 m/s2 without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible. Hint: The static friction between the tires and the road provides the forward force on the ford needed to accelerate it at 12 m/s2. 27. A block is launched up a frictionless incline, as shown below, with an initial speed of 5.5 m/s. What is the maximum displacement, d, of the block up the incline. Hint: The process of analyzing the forces acting upon objects on inclined planes will involve resolving the weight vector into two perpendicular forces. 28. The block shown below remains at rest. What is the friction force acting on the block? Hint: Use Newton’s Second Law then calculate using the angle. 29. What is the normal force acting on the block shown below? Hint: Use Newton’s second law then solve using the contact force. 30. A block slides down a frictionless, inclined plane that makes a 30o angle with the horizontal. Find the acceleration of this block. Hint: Solve for the acceleration by using Newton’s second law. FORMULAS USED IN THIS WORKSHEET Note: The following are general formulas, are not in any order, and have not necessarily been modified to fit each individual problem. Distance “d” can be expressed as “x” when horizontal or “y” when vertical. v 2 = v02 + 2a ( x − x0 ) v = d Δt v = v0 + at ∑ F = ma FT = m ( g + a ) ⎛ v 2 − v0 2 ⎞ F = ma = m ⎜ ⎟ ⎝ 2Δx ⎠ μs = D= FP mg vo2 2 g sin θ μs ≥ a g a= m= FT g+a F = mg sin θ a = − μk g F mg sin θ = = g sin θ m m y = y0 + v0 t + 12 at 2 a= FT − mg m FT = ma + mg = m ( a + g ) FP = Ffr = μk FN = μk mg μk = g sin θ − a g cos θ NAME: _______________________________ DATE: ___/___/___ PERIOD: _______ Honors Physics - Internet Assigned Problems Answer Sheet All worksheet answers must be submitted on this form and turned in on the due date. Your actual calculations must be stapled to the back of this form. Chapter: _______ Problems: ______________________________________ # FORMULA ANSWER UNIT # 1. 16. 2. 17. 3. 18. 4. 19. 5. 20 6. 21. 7. 22. 8. 23. 9. 24. 10. 25. 11. 26. 12. 27. 13. 28. 14. 29. 15. 30. FORMULA ANSWER UNIT