samira's questions.....plz help

[Samira]

Q1) aFind the horizontal force necessary to move a 10 kg object across a rough horizontal surfaceif the coefficient friction = 0.5

b) The force now acts at an angle of 30 degree to horizontal. Find the exact force necessary for the object to be on the point of moving.

k i figured out how to do part a nd the answer is 49; where the force should be greater than 49 .....but part b i cudn't get how to do.


Q2) 2 objects of masses 2.5 kg and 6 kg are joined by a string. A pulling force P is applied to the 6 Kg mass and the two begin moving across the surface with acceleration of 1.25m/s^2. The coefficient friction for each mass is 0.3. find T, tension n the string and P, if the pulling force makes an angke of 35 degree with horizantal.


thnx in advance

[kamil9876]

Vertical component:



  (1)

Point of moving implies that friction equals

Horizontal component:

(2)

use (1) and (2) to solve for F

[Samira]

thnx kamil

another qs
a girl is sitting at the top of  a 15 m long slide. The slide is inclined at an angle of 55 degree to vertical. she has a mass of 60 kg and the coefficient of friction between her nd the slide is 1/7

a) Find her acceleration
b) Find her speed at the bottom of slide
c) Find her momentum at the end

the ans r a)4.47 m/s^2      b)11.59m/s       c) 695.14 kg m/s

[TonyHem]

A)
15M long slide, 35 degrees to the horizontal, 60 kilograms.
So the force of being on a slope is MGsin(theta) = 60Gsin(35)
Friction opposes the direction of motion = normal * coefficient = 60GCOS(35)*(1/7)
This gives the resultant force when you minus them.
Then divide this by the mass for acceleration.
b)
Acceleration = 4.474236214
U = 0 V = ? X = 15
v^2 = u^=2+2ax
v = sqrt{2(a)(15)}
v = 11.585

c) p = mv m = 60 v = 11.585

[Samira]

thnx tonyhem

nother qs from heinemann specialist text

qs 20 pg 392 nd in qs 21 i just dnt get how to draw the forces

[TonyHem]

I'd help but I dont have the book :p

[Samira]

i wud wrote the qs bt i dnt noe how to draw the graph 4 yous thats the prob .... nyhow

nother qs
A Particle of mass 20 kg at rest is acted on by firce that decreases uniformly with the distance travelled. Thr force is imitially 100N and 30N after 20 min. what is The speed of the particle at this time?

ans: sqrt130

[TonyHem]

Initial acceleration = 5ms^-2
Final acceleration = 1.5ms^-2
Average acceleration = 3.25ms^-2

U = 0, V = ?, X = 20m, A = 3.25ms^-2
V = sqrt{130}

You made a typo (20 meters)( I did this question a week back)

[Samira]

oooppss my bad.....nywayz da qs was easyy i dnt noe y i cudnt get it
thnx

[kamil9876]

(1)


   (like a spring )




(2)

sub in v=0 when x=0 and you get c=0.

now when x=20, F=30. subbing into (1) you get a value for k. Then subbing into (2) you can solve for v.


This is a more general method. Averaging only worked because it was uniformly decreasing force and for non-uniform things average value is not just sum of initial and final divided by time.

[TonyHem]

I wanna see the above post :[, damn latex thing still isn't working.

[Samira]

Q) solve z^3 + (6+9i)z^2 + (15+36i)z + (46+73i)

Q)  solve z^2 + 2z- (sqrt3)i

[GerrySly]

Q)  solve z^2 + 2z- (sqrt3)i

Complete the square...



As a side, convert into form.



Now back to solving...