QUESTION 14** A ball of mass M is suspended vertically from the end of a string.

Question

QUESTION 14**

A ball of mass M is suspended vertically from the end of a string. The other end of the string is attached to the ceiling of an elevator, as shown in the figure. The elevator is initially moving downward at constant speed. Just before reaching the bottom floor, the elevator slows down. As it is slowing down, the tension in the string will be

(a)   greater than Mg.
(b)   equal to Mg.
(c)   less than Mg.

QUESTION 15*

This and the following question concern the same physical situation.

Block A of mass 2 kg and block B of mass 4 kg are attached to each other by a string. The two blocks are sitting on a horizontal frictionless surface. Another string is attached to block B and the whole system is pulled to the right so that both blocks accelerate together, as shown in the figure. If the tension T is 15 N, as shown in the figure, what is the tension T1 in the string connecting the two blocks?

(a)   1 N
(b)   3 N
(c)   5 N
(d)   12 N
(e)   15 N

QUESTION 16*

In the preceding question, suppose the order of the two blocks is reversed so that block A is in front and B is in back, with the same tension T = 15 N, as shown in the figure. Compared to the preceding problem, the tension T1 is now

(a)   greater.
(b)   less.
(c)   the same.

QUESTION 17**

This and the following question concern the same physical situation.

A block of mass M = 6 kg slides down an incline at constant velocity. The incline makes an angle of 40

Explanation / Answer

14.)pseudo force downward so greater than Mg (A)

15.) acc= 15/2+4 = 2.5

so tension = mass * acc = 2.5 * 2 = 5 (C)

16.)new tension = 4* 2.5 = 10N greater (A)

17.)given constant veocity

friction = mg*sin(40) = mew * normal reaction = mew * mg*cos(40)

mew = tan(40) = 0.839 (B)

18.)False (B)

19.)deccleration = mew * g = vel/time = 7.5/2.0

mew=0.3822 (E)

20.)4

friction ,applied force,normal reaction , weight

(C)

21.) weight = friction

mew* 10 = m*g

m=0.601 (D)

22.)tension will be only due to 30kg weight

T= 30* g= 294.3 (E)

23.)writing equation of motion and solving we get T=183.93 lesser (B)

24.)Lesser (B)

25.)seed inversely proportional to aqrt of distance so

(B) the one closer to the center of the earth

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