All Mechanics DOUBTS HERE!!!

[Phosu]

see we kno total distance is 125m
125= 78.6+x
find x
which is 48.6m

so she travels 48.6m for t seconds

now to find t
48.6 = 4 (t+7)-7
48.6 =4t
t=48.6/4 = 11.6s

[cooldude]

A load of weight 7kN is being raised from the rest with constant acceleration by a cable. After the load has been raised 20 metres, the cable suddenly becomes slack. The load continues to move upwards for a distance of 4 metres before coming to an instantaneous rest. Assume no air resistance, find the tension in the cable before it became slack.

this is for the first part of the journey, let the tension in the string be T and the velocity of the load after it has travelled 20m--->

T-7000=700a
v^2=u^2+2as
v^2=2*20*((T-7000)/700)

for the second part of the journey (when the string is slack), note in the second line, v^2 is referred to as the velocity found above-->
v^2=u^2+2as
0=v^2+2*-10*4
v^2=80
now substitute the value of v^2 in the above equation-->
80=40((T-7000)/700)
T=1400+7000=8400 N

[halosh92]

see we kno total distance is 125m
125= 78.6+x
find x
which is 48.6m

so she travels 48.6m for t seconds

now to find t
48.6 = 4 (t+7)-7
48.6 =4t
t=48.6/4 = 11.6s

thxxx aloooottt
all clear 

[Ghost Of Highbury]

this is for the first part of the journey, let the tension in the string be T and the velocity of the load after it has travelled 20m--->

T-7000=700a
v^2=u^2+2as
v^2=2*20*((T-7000)/700)

for the second part of the journey (when the string is slack), note in the second line, v^2 is referred to as the velocity found above-->
v^2=u^2+2as
0=v^2+2*-10*4
v^2=80
now substitute the value of v^2 in the above equation-->
80=40((T-7000)/700)
T=1400+7000=8400 N

ok got it , thanks a lot

[wakemeup]

Man, edexcel jan 2009 is insane. Q5a. Why is the friction in the upward direction? Q5b. I have no idea what's going on here. Q7b and c. Again, no idea at all. Please help...

http://www.xtremepapers.net/Edexcel/Mathematics/2009%20Jan/2009%20Jan%20QPM1.pdf

[Darkstar3000]

At time t=0 at particle is projected vertically upwards with speed u m/s from a point 10m above the ground. At time T seconds, the particle hits the ground with speed 17.5 m/s . Find

a) The value of u (i can find this)

b) The value of T

part b is my problem because in the marking scheme they used V=u + at and took the initial speed (u)  as negative, why do we take it as a negative because in my calculation i wrote it like this: 17.5=10.5 - 9.8t

this is from May 2008 Q2

[cooldude]

darkstar3000--->

(m doin the a part as i need the value of u)

a) highest point reached by the object-->
v^2=u^2+2as
0=u^2+(2*-10s)
u^2=20s
b) v^2=u^2+2as
17.5^2=0+(2*10*(10+s)
s=5.3125
u=10.31 m/s

b) for the upward journey time--->
v=u+at
0=10.31-10t
t=1.031 sec
for the downward journey time-->
v=u+at
17.5=0+10t
t=1.75 sec
therefore T=1.75+1.031=2.78 sec

and now ur qs, in these type of qs its best to divide ur answer into two parts (the upward part and the downward part), uve assumed that g is negative for the whole journey but that is only true for the upward journey, thus u will not get the correct answer

[Darkstar3000]

darkstar3000--->

(m doin the a part as i need the value of u)

a) highest point reached by the object-->
v^2=u^2+2as
0=u^2+(2*-10s)
u^2=20s
b) v^2=u^2+2as
17.5^2=0+(2*10*(10+s)
s=5.3125
u=10.31 m/s

b) for the upward journey time--->
v=u+at
0=10.31-10t
t=1.031 sec
for the downward journey time-->
v=u+at
17.5=0+10t
t=1.75 sec
therefore T=1.75+1.031=2.78 sec

and now ur qs, in these type of qs its best to divide ur answer into two parts (the upward part and the downward part), uve assumed that g is negative for the whole journey but that is only true for the upward journey, thus u will not get the correct answer

thank you very much

[wakemeup]

Man, edexcel jan 2009 is insane. Q5a. Why is the friction in the upward direction? Q5b. I have no idea what's going on here. Q7b and c. Again, no idea at all. Please help...

http://www.xtremepapers.net/Edexcel/Mathematics/2009%20Jan/2009%20Jan%20QPM1.pdf

                                                                                                                                                                                             Please, people. Need answers urgently.

[Igcseboy]

Hello every1 plzz i need answers to the  following questions

Jan 2009 qustion 5 b i and ii

May 2008 6 b

May 2008 5 a and b

Jan 2008 7 c

Jan 2008 6 d

Any1 able to answer these plzzz do so

May God bless us All

[Saladin]

Hello every1 plzz i need answers to the  following questions

Jan 2009 qustion 5 b i and ii

May 2008 6 b

May 2008 5 a and b

Jan 2008 7 c

Jan 2008 6 d

Any1 able to answer these plzzz do so

May God bless us All

2moro!

[halosh92]

Q3
plzzz someone explainnn
thx a bunch URGENTTTT

[astarmathsandphysics]

,moments about A -2*2+3*2+4*0+5*0=2Nm anticlockwise

[astarmathsandphysics]

This is 7c
Resolving vertically-6.75g-6.75gcos36.87=-12.15g
resolving horizontally -6.75gsinn36.87=-4.5g
R=sqrt((-12.15g)^2+(-4.5g)^2)=12.96g

[Saladin]

This is 7c
Resolving vertically-6.75g-6.75gcos36.87=-12.15g
resolving horizontally -6.75gsinn36.87=-4.5g
R=sqrt((-12.15g)^2+(-4.5g)^2)=12.96g

I owe you one.