Mechanics

[simba]

Q: A of mass 4 kg is pulled from rest to a speed of 4.5m/s in time of 3 seconds on a rough horizontal surface by a force of 20N which makes an angle of 10 degrees with the horizontal.Find the coefficient of friction....



will sm1 help !!!

[astarmathsandphysics]

Res vertically
R+20sin10-4g=0 so R=4g-20sin10
res horizontally 20cos10-uR=ma=4*(4.5/3)=6 so uR=20cos10-6
u=(20cos10-6)/(4g-20sin10)

[Phosu]

doubt

edexcel m1 jan 03 ques 7b

marking scheme says
-1.5=22.4t-4.9t^2

what i dont understand is why is -1.5 being used as the distance...?

[nid404]

Could you post the paper cuz i don't have it...

my guess is the distance travelled may be in the opp dir....

[astarmathsandphysics]

Without having read the question I agree. Did it land below the Start point?

[Phosu]

yes it does land below the starting point...

A ball is projected vertically upwards with a speed u m s?1 from a point A which is 1.5 m above the ground. The ball moves freely under gravity until it reaches the ground. The greatest height attained by the ball is 25.6 m above A.
The ball reaches the ground T seconds after it has been projected from A.
(a) Show that u = 22.4
(b) Find, to 2 decimal places, the value of T.

[nid404]

a)
v²=u²+2as
velocity and max height=0
0=u²- 2gX25.6  ( - because ball decelerates when it's moving upwards)
u²=2gX25.6
u=22.4m/s

b) v=u+at for the first part(upward motion)
0=22.4-10t
t=2.24s to reach max height

comin down (25.6+1.5)m
s=1/2 Xat2
27.1=1/2X10 t2
t=2.32s
total time= 2.24+2.32=2.56s

[simba]

i hv a question :   


A load of mass 50kg stands on the horizontal floor of a goods lift which is accelerating vertically downwards at 2 m/s square. Calculate the magnitude of the contact force exerted on the load by the floor.

[astarmathsandphysics]

50(9.8-2)=390N

[simba]

the answer is 400, nt 390 .......i saw it bt i dnt knw how it is done

[astarmathsandphysics]

I used g=9.8 I think they used g=10