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1. nT5B-RT16: ROPES PULLING BOXES—ROPE TENSION
Shown below are boxes that are being pulled by ropes along frictionless surfaces, accelerating toward the left. The accelerations of the boxes are the same, and the masses of the boxes are indicated in each figure.
Rank the magnitude of the tension in these ropes.
Largest 1________ 2________ 3________ 4________ 5________ 6________ Smallest
OR, All of these ropes have the same tension. ___
OR, We cannot determine the ranking for the tensions. ___
1. NT5G-RT58: SYSTEM OF ACCELERATING BLOCKS—TENSION
A system of two blocks connected by a massless string is pulled so that the system is accelerating. There is no friction between the blocks and the surfaces they move along. Values for six variations (labeled A-F) of this system that have different masses and accelerations are given in the table.
P I II Case
A
AccelerationB
C
D
E
F
18 kg 10 kg 15 kg 10 kg 15 kg 15 kg 18 kg 15 kg 15 kg 15 kg 18 kg 10 kg
2 m/s2 2 m/s2 2 m/s2 2 m/s2 3 m/s2 3 m/s2 MI MII
a
a
Rank these situations on the basis of the tension in the string at point P.
2. NT5B-RT6: TUGBOAT PUSHING BARGES—FORCE TUGBOAT EXERTS ON FIRST BARGE
Each of the six figures below shows a system consisting of a tugboat pushing two barges labeled 1 and 2. The masses of the tugboats and the barges along with the accelerations of the systems are given for each case. The positive direction is to the right. Ignore the effects of fluid friction.
2. NT5B-RT9: TIME-VARYING FORCE ON A CART—ACCELERATION
The graph below shows the net force in the x-direction acting on a cart as a function of time.
A Time
Force in x–direction
0
F E
D C
B
Rank the acceleration of the cart at the labeled times.
3. NT5G-WBT90: NEWTON’S SECOND LAW EQUATIONS—PHYSICAL SITUATION
Given below are two equations resulting from the application of Newton’s Laws to a system of two objects:
Mg – T = Ma T + µmg = ma
3. NT5H-CT95: FORCE ON BOX MOVING OVER HORIZONTAL SURFACE—FRICTIONAL FORCE ON BOX In both cases below, a moving 50 N box has a force on it of 40 N that makes an angle of 30º with the horizontal. The coefficient of static friction between the box and the rough surface is 0.6 and the coefficient of kinetic friction is 0.4.
A
B
50 N
40 N 30°
50 N 30°40 N
Will the frictional force exerted on the box by the rough surface in Case A be greater than, less than, or equal to thefrictional force on the box by the rough surface in Case B?
4. NT5G-WWT67: HANGING STONE AND SLIDING BOX—FREE-BODY DIAGRAMS
A massless rope connects a box on a horizontal surface and a hanging stone as shown below. The rope passes over a massless, frictionless pulley. The box is given a quick tap so that it slides to the right along the horizontal surface. The figure below shows the block after it has been pushed while it is still moving to the right. The mass of the hanging stone is larger than the mass of the box. There is friction between the box and the horizontal surface. Free-body diagrams that a student has drawn to scale for the box and for the hanging stone are shown.
240 g v = 2 m/s
Fon stone by rope
Fon stone by earth
Fon box by surface
Fon box by earth Fon box by rope
Fon box by surface
5. NT5G-CCT76: STRINGS CONNECTED TO A RING—TENSION IN STRINGS The three strings in the arrangement shown are massless, and the mass of the ring can also be ignored. Strings 1 and 2 are the same length, and each makes an angle of 60° with the vertical. Four students make the following contentions:
Arlene: “The tension in strings 1 and 2 will each be half the weight of the suspended mass. Each holds up half of the weight.”
Bablu: “The tension in strings 1 and 2 will each be greater than half the weight of the suspended mass. These strings have to fight each other as well as hold up the suspended mass.” Cain: “I think you have to calculate to decide. The tensions in strings 1 and 2 will depend on the
sine of the angle, and since sine and cosine functions can never be greater than one the tension will be some fraction of the mass.”
Danyl: “We can’t compare the tension in strings 1 and 2 to the weight of the suspended mass unless
we know the length of String 3. If string 3 is long enough, then to compensate the vectors for strings 1 and 2 will have to be really large.”
Which, if any, of these students do you agree with?
5. NT5G-CCT78: UNEQUAL LENGTH STRINGS CONNECTED TO A RING—TENSION IN STRINGS The three strings in the arrangement shown are massless, and the mass of the
ring can also be ignored. String 1 is 1 meter long, and string 2 is 2 meters long. Four students make the following contentions:
Aba: “The tension in strings 1 and 2 will each be half the weight of the suspended mass. They each have to hold up half of the weight.” Belita: “The tension in string 1 will be greater than the tension in string 2,
because the string is shorter. The tension is more concentrated in a shorter string.”
Coco: “String 1 is at twice the angle but has only half the length. These effects compensate, so the tension in string 1 is the same as the tension in string 2.”
David: “The tension in string 2 will be greater than the tension in string 1 because this string is at a smaller angle to the vertical. When you add up the forces it will be the long side of the triangle.”
Which, if any, of these students do you agree with?
Aba_____ Belita _____ Coco _____ David _____ None of them______ Explain.
Ring String 3 String 1
String 2 30°
60°
