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Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: simba on February 23, 2010, 10:29:46 am

Title: Mechanics
Post by: simba on February 23, 2010, 10:29:46 am

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 !!!

Title: Re: Mechanics
Post by: astarmathsandphysics on February 23, 2010, 10:58:52 am

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)

Title: Re: Mechanics
Post by: Phosu on February 26, 2010, 08:43:32 am

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...?

Title: Re: Mechanics
Post by: nid404 on February 26, 2010, 09:45:27 am

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

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

Title: Re: Mechanics
Post by: astarmathsandphysics on February 26, 2010, 10:29:29 am

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

Title: Re: Mechanics
Post by: Phosu on February 26, 2010, 10:49:32 am

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.

Title: Re: Mechanics
Post by: nid404 on February 26, 2010, 11:14:52 am

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

Title: Re: Mechanics
Post by: simba on March 02, 2010, 09:59:36 am

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.

Title: Re: Mechanics
Post by: astarmathsandphysics on March 02, 2010, 10:13:26 am

50(9.8-2)=390N

Title: Re: Mechanics
Post by: simba on March 02, 2010, 10:25:43 am

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

Title: Re: Mechanics
Post by: astarmathsandphysics on March 02, 2010, 11:04:39 am

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