Fall Final Review WKS: WORD PROBLEMS d Average Speed v t 1. A rock is dropped from the top of a tall cliff 9 meters above the ground. The ball falls freely and hits the ground 1.5 seconds later. What is the average speed of the rock? d Average Speed v t 1. A rock is dropped from the top of a tall cliff 9 meters above the ground. The ball falls freely and hits the ground 1.5 seconds later. What is the average speed of the rock? d v t 9m v 1 .5 s v 6m / s 2. A car travels at a constant speed of 20 m/s for 5 seconds. How far did it travel? 2. A car travels at a constant speed of 20 m/s for 5 seconds. How far did it travel? d v t Rearrange to get d vt d 20m / s 5s d 100 m 3. A car travels 45 miles in 1 hour. What was the car’s average speed? 3. A car travels 45 miles in 1 hour. What was the car’s average speed? d v t 45 miles v 1hr v 45miles / hr Uniformly Accelerated Motion v v0 gt 4. You drop a rock off the top of a tall building and it hits the ground 4 seconds later. What is the velocity of the rock when it hits the ground? Uniformly Accelerated Motion v v0 gt 4. You drop a rock off the top of a tall building and it hits the ground 4 seconds later. What is the velocity of the rock when it hits the ground? v v0 gt time final velocity intial velocity g =10 Uniformly Accelerated Motion v v0 gt 4. You drop a rock off the top of a tall building and it hits the ground 4 seconds later. What is the velocity of the rock when it hits the ground? V0=0 v v0 gt t=4s v 0 104s V=? v 40m / s Newton’s Second Law F ma 5. A 2.5 N object is sitting at rest on the table. What is the force of the table acting on the object? What is the net force? Newton’s Second Law F ma 5. A 2.5 N object is sitting at rest on the table. What is the force of the table acting on the object? What is the net force? Fg = -2.5N 2.5N Support force -2.5N Force or gravity (weight) 2.5N + (-2.5N) = 0N 6. You are ice skating with a friend and he pushes you across the frictionless ice with a force of 5 Newtons. If your mass is 50 kg, what will you acceleration be? 6. You are ice skating with a friend and he pushes you across the frictionless ice with a force of 5 Newtons. If your mass is 50 kg, what will you acceleration be? F a m 5N a 50kg a 0.1 m s 2 7. If we disregard friction, what force would be required to accelerate an 10 kg object at 4.0 m/s2 ? 7. If we disregard friction, what force would be required to accelerate an 10 kg object at 4.0 m/s2 ? F ma F 10kg 4.0m / s F 40 N 2 Force due to gravity Fg mg 8. What is the weight of a 10 kg mass on Earth? Force due to gravity Fg mg 8. What is the weight of a 10 kg mass on Earth? Fg = weight Fg mg Fg 10kg 10 Fg 100N 9. A rock has a weight of 98 N. What is the mass? 9. A rock has a weight of 98 N. What is the mass? Fg = weight Fg mg 98N m 10 Rearrange and get m m 9.8kg Fg g 10. Which has a greater mass. A 20 N object on the moon, or a 20N object on the Earth? (why) 10. Which has a greater mass. A 20 N object on the moon, or a 20N object on the Earth? (why) 20N object = Fg of the object mmoon Fg mearth Fg g g mmoon 20 N 12 kg 1.66 mearth 20 N 2kg 10 Remember that value of g varies on the moon and the Earth. Moon is only 1/6th that on Earth 11. You are pulling a 50 kg crate with a force of 5 N. If there is 2 N of friction, what is the acceleration of the crate? 11. You are pulling a 50 kg crate with a force of 5 N. If there is 2 N of friction, what is the acceleration of the crate? First calculate the net force- F a m 3N a 50kg 5N – 2N = 3N (to the right) a 0.06m / s 2 Work W Fd Power W P t 12. How much work is done pushing a 10N crate a distance of 10 meters? Work W Fd Power W P t 12. How much work is done pushing a 10N crate a distance of 10 meters? W Fd W 100 J W 10N 10m Remember that the units for both work and energy are Joules Kinetic Energy 1 2 E mv 2 Gravitational Potential Energy E m gh 14. What is the kinetic energy of a 5.0 kg rock moving at 2 m/s? Kinetic Energy 1 2 E mv 2 14. What is the kinetic energy of a 5.0 kg rock moving at 2 m/s? 1 2 E 5.0kg 2m / s 2 1 2 E mv 2 1 E 5.0 4 2 Remember to start by squaring velocity E 10 J 15. A 5 kg rock falls off the top of a 5 meter tall roof. What is the approximate kinetic energy of the rock just before it hits the ground? You are not given the velocity, so you will need to use the equation for PE understanding that it will all be h=5m transformed into KE as it falls E m gh E 250 J E 5kg 105m It starts with 250J of PE that is converted to 250J of KE 16. What is the kinetic energy of a 20 Newton object that is moving at 5 m/s? 16. What is the kinetic energy of a 20 Newton object that is moving at 5 m/s? You will first need to find out what the mass of the object is using equation for Fg=mg Fg mg rearrange 1 2 E mv 2 m Fg g 20 N m 2kg 10 1 2 E 2kg 5m / s 2 E 25 J 17. What is the gravitational potential energy given to a 3.0 kg object that is lifted 2.0 meters off of the ground? 17. What is the gravitational potential energy given to a 3.0 kg object that is lifted 2.0 meters off of the ground? E m gh E 3kg 102.0m E 60 J Momentum p mv 18. What is the momentum of a 5.0 kg ball moving at 6 m/s? 18. What is the momentum of a 5.0 kg ball moving at 6 m/s? p mv p 5.0kg 6m / s p 30kg m / s 19. A 75 kg skier is moving down a hill with a momentum of 100 kg·m/s. What is the velocity of the skier? 19. A 75 kg skier is moving down a hill with a momentum of 100 kg·m/s. What is the velocity of the skier? p mv Rearrange and solve for v p v m p v m 100kg m / s v 75kg v 1.33m / s 20. What is the mass of an object that has a momentum of 60 kg·m/s and a velocity of 10 m/s? 20. What is the mass of an object that has a momentum of 60 kg·m/s and a velocity of 10 m/s? p mv p m v Rearrange and solve for m p m v 60kg m / s m 10m / s m 6kg Collision in One Dimension 21. What is the final velocity of the coupled railroad cars after the inelastic collision? Collision in One Dimension 21. What is the final velocity of the coupled railroad cars after the inelastic collision? To solve these problems you need to remember two things: •pbefore = pafter •Use equation p=mv Collision in One Dimension 21. What is the final velocity of the coupled railroad cars after the inelastic collision? Momentum before = pcar1 mv pcar1 57 35 pcar 2 mv pcar 2 20 0 ptotal 35 0 35 Remember total pbefore = pafter pafter mv 35 5 2v pafter =35 35 vafter 5m / s 7 22. What is the mass of the second (smaller sized) box? 22. What is the mass of the second (smaller sized) box? To solve these problems you need to remember two things: •pbefore = pafter •Use equation p=mv 22. What is the mass of the second (smaller sized) box? Momentum before = pblock1 mv pblock1 26 12 pblock 2 mv pblock 2 20 0 ptotal 12 0 12 Remember total pbefore = pafter pafter mv 12 m3 pafter =12 12 mtotal 4kg 3 Mass of second block: 4kg – 2kg = 2kg 23. What is the velocity of the gray ball after the elastic collision below? Assume each ball has the same mass. 23. What is the velocity of the gray ball after the elastic collision below? Assume each ball has the same mass. 4 m/s In order for momentum to be conserved, the total momentum that exists before will equal the total after the collision. If each ball has the same mass, the velocity of the ball after collision must equal the velocity of the ball before