Module 4 Recitation 2 Concept Problems Concep Test

Module 4, Recitation 2 Concept Problems

Concep. Test Force and Two Masses A force F acts on mass m 1 giving acceleration a 1. The same force acts on a different mass m 2 giving acceleration a 2 = 2 a 1. If m 1 and m 2 are glued together and the same force F acts on this combination, what is the resulting acceleration? F F F m 1 a 1 m 2 m 1 a 2 = 2 a 1 a 3 1) 3/4 a 1 2) 3/2 a 1 3) 1/2 a 1 4) 4/3 a 1 5) 2/3 a 1

Concep. Test Force and Two Masses A force F acts on mass m 1 giving acceleration a 1. The same force acts on a different mass m 2 giving acceleration a 2 = 2 a 1. If m 1 and m 2 are glued together and the same force F acts on this combination, what is the resulting acceleration? F m 1 a 1 F m 2 F = m 2 a 2 = (1/2 m 1 )(2 a 1 ) m 2 m 1 2) 3/2 a 1 3) 1/2 a 1 4) 4/3 a 1 5) 2/3 a 1 F = m 1 a 2 = 2 a 1 F 1) 3/4 a 1 Mass m 2 must be (1/2)m 1 because its acceleration was 2 a 1 with the same force. Adding the two masses together gives (3/2)m 1, a 3 leading to an acceleration of (2/3)a 1 for the same applied force. F = (3/2)m 1 a 3 => a 3 = (2/3) a 1

Concep. Test Contact Force I Box 1 and 2 are on a horizontal frictionless surface. If you push with force F on either the heavy box (m 1) or the light box (m 2), in which of the two cases below is the contact force between the two boxes larger? 1) case A 2) case B 3) same in both cases A m 2 F m 1 B m 2 m 1 F

Concep. Test Contact Force I Box 1 and 2 are on a horizontal frictionless surface. If you push with force F on either the heavy box (m 1) or the light box (m 2), in which of the two cases below is the contact force between the two boxes larger? The acceleration of both masses together is 1) case A 2) case B 3) same in both cases A the same in either case. But the contact force is the only force that accelerates m 1 m 2 F m 1 in case A (or m 2 in case B). Since m 1 is the B larger mass, mass it requires the larger contact force to achieve the same acceleration. m 2 m 1 F

Concep. Test Contact Force II Two blocks of masses 2 m and m 1) 2 F are in contact on a horizontal 2) F frictionless surface. If a force F 3) 1/2 F is applied to mass 2 m, what is 4) 1/3 F the force on mass m ? 5) 1/4 F F 2 m m

Concep. Test Contact Force II Two blocks of masses 2 m and m 1) 2 F are in contact on a horizontal 2) F frictionless surface. If a force F 3) 1/2 F is applied to mass 2 m, what is 4) 1/3 F the force on mass m ? 5) 1/4 F The force F leads to a specific acceleration of the entire system. In order for mass m to accelerate at the same rate, the force on it must be smaller! How small? ? Let’s see. . . F 2 m m

Concep. Test Tension I You tie a rope to a tree and you pull on the rope with a force of 100 N. What is the tension in the rope? 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N

Concep. Test Tension I You tie a rope to a tree and you pull on the rope with a force of 100 N. What is the tension in the rope? 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N The tension in the rope is the force that the rope “feels” across any section of it (or that you would feel if you replaced a piece of the rope). Since you are pulling with a force of 100 N, that is the tension in the rope.

Concep. Test Tension II Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope? 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N

Concep. Test Tension II Two tug-of-war opponents each pull with a force of 100 N on opposite ends of a rope. What is the tension in the rope? 1) 0 N 2) 50 N 3) 100 N 4) 150 N 5) 200 N This is literally the identical situation to the previous question. The tension is not 200 N !! Whether the other end of the rope is pulled by a person, or pulled by a tree, the tension in the rope is still 100 N !!

Concep. Test You and a friend can each pull with a force of 20 N. If you want to rip a rope in half, what is the best way? Tension III 1) you and your friend each pull on opposite ends of the rope 2) tie the rope to a tree, and you both pull from the same end 3) it doesn’t matter -- both of the above are equivalent 4) get a large dog to bite the rope

Concep. Test You and a friend can each pull with a force of 20 N. If you want to rip a rope in half, what is the best way? Tension III 1) you and your friend each pull on opposite ends of the rope 2) tie the rope to a tree, and you both pull from the same end 3) it doesn’t matter -- both of the above are equivalent 4) get a large dog to bite the rope Take advantage of the fact that the tree can pull with almost any force (until it falls down, that is!). You and your friend should team up on one end, and let the tree make the effort on the other end.

Concep. Test Three Blocks Three blocks of mass 3 m, 2 m, and m are connected by strings and 1) T 1 > T 2 > T 3 2) T 1 < T 2 < T 3 pulled with constant acceleration 3) T 1 = T 2 = T 3 a. What is the relationship 4) all tensions are zero between the tension in each of 5) tensions are random the strings? a 3 m T 3 2 m T 2 m T 1

Concep. Test Three Blocks Three blocks of mass 3 m, 2 m, and m are connected by strings and 1) T 1 > T 2 > T 3 2) T 1 < T 2 < T 3 pulled with constant acceleration 3) T 1 = T 2 = T 3 a. What is the relationship 4) all tensions are zero between the tension in each of 5) tensions are random the strings? T 1 pulls the whole set of blocks along, so it a must be the largest T 2 pulls the last two masses, but T 3 only pulls the last mass. 3 m T 3 2 m T 2 m T 1 Follow-up: What is T 1 in terms of m and a?

Concep. Test Over the Edge 1) case 1 In which case does block m experience a larger acceleration? In 2) (1) there is a 10 kg mass hanging 3) from a rope and falling. In (2) a 4) hand is providing a constant 5) downward force of 98 N. Assume acceleration is zero both cases are the same depends on value of m case 2 massless ropes. m m 10 kg a a F = 98 N Case (1) Case (2)

Concep. Test Over the Edge In which case does block m experience 1) case 1 a larger acceleration? In (1) there is a 2) 10 kg mass hanging from a rope and 3) falling. In (2) a hand is providing a 4) constant downward force of 98 N. 5) Assume massless ropes. In (2) the tension is 98 N because the block 10 kg a rest would the tension be equal to 98 N. depends on value of m case 2 m a F = 98 N is accelerating down Only if the block were at both cases are the same m due to the hand. In (1) the tension is less than acceleration is zero Case (1) Case (2)

Concep. Test Going Up I A block of mass m rests on the 1) N > mg floor of an elevator that is moving 2) N = mg upward at constant speed. What 3) N < mg (but not zero) is the relationship between the force due to gravity and the normal force on the block? 4) N = 0 5) depends on the size of the elevator v m

Concep. Test Going Up I A block of mass m rests on the 1) N > mg floor of an elevator that is moving 2) N = mg upward at constant speed. What is 3) N < mg (but not zero) the relationship between the force due to gravity and the normal force on the block? 4) N = 0 5) depends on the size of the elevator The block is moving at constant speed, so it must have no net force on it. The forces v on it are N (up) and mg (down), so N = mg, just like the block at rest on a table. m

Concep. Test Going Up II A block of mass m rests on the 1) N > mg floor of an elevator that is 2) N = mg accelerating upward. What is 3) N < mg (but not zero) the relationship between the 4) N = 0 force due to gravity and the 5) depends on the size of the elevator normal force on the block? a m

Concep. Test Going Up II A block of mass m rests on the 1) N > mg floor of an elevator that is 2) N = mg accelerating upward. What is 3) N < mg (but not zero) the relationship between the force due to gravity and the normal force on the block? 4) N = 0 5) depends on the size of the elevator The block is accelerating upward, so it must have a net upward force The forces on it are N (up) and mg (down), so N must be greater than mg in order to give the net upward force! force Follow-up: What is the normal force if the elevator is in free fall downward? N m a>0 mg S F = N – mg = ma > 0 N > mg