I need explanations! Not only answers! QUESTION 12* This and the following quest

Question

I need explanations!

Not only answers!

QUESTION 12*

This and the following question concern the same physical situation.

A pilot must fly her plane due north to reach her destination. ?he plane flies at 300 km/h in still air. ?f a wind is blowing from the west at 60 km/h, what is the direction the pilot must head her plane in order to fly straight north?

(a)   6.1

I need explanations! Not only answers! QUESTION 12* This and the following question concern the same physical situation. A pilot must fly her plane due north to reach her destination. ?he plane flies at 300 km/h in still air. ?f a wind is blowing from the west at 60 km/h, what is the direction the pilot must head her plane in order to fly straight north? (a) 6.1° east of north (b) 12.2° east of north (c) 14.2° west of north (d) due west (e) 11.5° west of north QUESTION 13*** Assuming she is heading in the direction found in the previous problem, the speed of the plane relative to the ground is (a) greater than 300 km/h. (b) 300 km/h. (c) less than 300 km/h. QUESTION 14* This and the following two questions concern the same physical situation. A block of mass m1 = 2 kg is on a frictionless horizontal surface. ? light string that passes over frictionless and massless pulley pulls it. ?he other end of the string is connected to a block of mass m2 = 3 kg.½ What is the acceleration of the blocks? (a) 2.94 m/s2 (b) 2.0 m/s2 (c) 11.7 6m/s2 (d) 4.0 m/s2 (e) 5.88 m/s2 QUESTION 15** What is the tension in the string? (a) greater than 29 N (b) less than 29 N (c) 29 N QUESTION 16* Let the tension in the preceding problem be T.½ Suppose now that there is sliding friction between the 2 kg block and the horizontal surface.½ The tension in the string will be (a) equal to T. (b) larger than T. (c) smaller than T. QUESTION 17* This and the following question concern the same physical situation. A small disk of mass m = 0.2 kg is tied to the end of a light string and twirled at constant speed v = 3 m/s in a circle of radius R = 4 m on a horizontal frictionless surface. What is the tension in the string? (a) 0.45 N (b) 0.6 N (c) 1.8 N (d) 2.7 N (e) 0.23 N QUESTION 18* Let F be the answer to the previous problem.½ Suppose the radius of the circle is changed to 8 m.½ The tension in the string is now (a) greater than F. (b) equal to F. (c) less than F. QUESTION 19* A 50-kg crate rests on the bed of a truck which is accelerating to the right with a = 3.9 m/s2.½ What is the minimum coefficient of static friction between the surface of the truck and the crate such that the crate does not slip? (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4 (e) 0.5 QUESTION 20* This and the following question concern the same physical situation. A skier glides down a frictionless slope that is inclined at q = 13½ to the horizontal. What is the acceleration of the skier? (a) 0.58 m/s2 (b) 2.2 m/s2 (c) 9.81 m/s2 (d) 0.15 m/s2 (e) 0.29 m/s2 QUESTION 21** Suppose instead that the coefficient of kinetic friction between the skis and the snow is 0.20.½ What is the acceleration of the skier? (a) 0.58 m/s2 (b) 2.2 m/s2 (c) 9.81 m/s2 (d) 0.15 m/s2 (e) 0.29 m/s2 QUESTION 22* A box with an initial speed of 15 m/s slides along a surface where the coefficient of sliding friction is 0.45.½ How long does it take for the block to come to rest? (a) 1.5 s (b) 3.4 s (c) 6.8 s (d) 8.5 s (e) 4.7 s QUESTION 23** A box of mass M rests on the floor of an elevator which is going up the Sears tower.½ As it approaches the top, the elevator starts to slow down.½ At this point, the normal force exerted by the floor on the box is (a) less than Mg. (b) equal to Mg. (c) greater than Mg.

Explanation / Answer

12. d

Need to balance velocitie in east-west direction. Wind will push towards east hence plane's velocity must in north-west direction. 300 cos(x) = 60 (balancing components of velocities)

13. c

Now speed relative to ground is combined effect of plane's engine and wind, hence only the north component of plane's velocity which will be lesser than 300 (=300 cosx)

14. e

Writing Eq of Newton's Law of Motion:

3g - T = 3a

T = 2a

Hence, a = 3g/5

15. b

T = 2a = 6g/5

16. c

If friction is there it will can't exceed Tension (as it will lead to motion in reverse direction which is not possible). It will not be equal to Tension bcz max friction (say, for u = 1) will be umg = 2g, which is not enough to keep the 3 kg block at rest.

17. a

F = (mv2)/r = (0.2*9) / 4

18. c

Acc to above formula if r increases, F should decrease.

19. d

Friction force = mass * acceleration of truck

umin*(mg) = ma

umin = a/g

20. b

Take component of weight (gravitational force) along the slide.

Falong = mg sin(theta) = mgsin(13o)

acc = F/m

21. e

Falong = mg sin(theta) - Frictional Force = mg sin(theta) - umg cos (theta)

a = Falong/m

22. b

v = v0 + at

As, v = 0,

t = -v0/a = -15 / (-ug)

23. c

Greater than Mg bcz not only it has to push its weight upwards but also account for 'a' accelearation upwards.

Fnet = mg + ma (where 'a' is acceleration of lift upwards)

Cheers

PS - Feel free to ask doubt in any part.

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