At a certain point the speed of the bucket begins to change. The bucket now has an upward constant acceleration of magnitude 3 m/s2. What is the tension in the rope now?

a)  T = 58.8 N ( with acceleration a= 0)

b)  T= 76.8 N (up ward motion with acc a=3m/s)

c)  T = 40.8 N (downward motion with acc -3m/s)

Explanation:

The question exact statement is :

A 6kg bucket of water is being pulled straight up by a string at a constant speed. I determined that the tension on the string was F = ma

F = (6kg * 9.8 m/s2) , F = 58.8 N Now its asking At a certain point the speed of the bucket begins to change. The bucket now has an upward constant acceleration of magnitude 3 m/s2. What is the tension in the rope now? The correct answer was "about 78N" then Now assume that the bucket has a downward acceleration, with a constant acceleration of magnitude3 m/s2. Now what is the tension in the rope?

Solution:

a) Motion with constant speed

Mass of bucket= m 6 kg

Weight force of body =mg= 6×9.8=58.8 N

Speed is constant so a=0 m/sec2

So net force is tension

T in the string

and

weight  mg of the body!

Net force is;

T-mg =ma (When bucket move up tension will greater than weight - so we subtract small force from bigger force)

==> T-58.8=6×0=0

==> T = 58.8 N

b) When bucket rise up with acceleration a

Here again;

m= 6 kg

g= 9.8 m/sec2

and a= 3 m/sec2

Again net force is

Tension=T

Weight = mg = 538.8N

So, net force is

T-mg = ma

==> T-58.8=6×3

==> T = 18+58.8=76.8 N

c) When bucket comes down with acceleration a

Again

m= 6 kg

g= 9.8 m/sec2

and a=  - 3 m/sec2 ( because now acceleration is in opposite direction)

Again net force is

Tension=T

Weight = mg = 58.8N

So, net force is

T-mg = ma

==> T-58.8=6 × (-3)

==> T= -18 +58.8

==> T = 40.8 N

28. evolution

29. environment, survive, offspring  

explanation:

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)