Mechanics 42

[Saad Sarfraz]

SO TRUE!

Okay could u help me on this the curves. i drew them right till 2 sec mark. then i messed up each. any prediction how many marks ill loose?

 

[Gooners]

oh, sry. the answers were 945 kj and 987 kj. No, you dnt have to consider weight.

In the part where velocity was CONSTANT, (i) u had to take the componemt of weight and add it to friction thn multiply by 400m!
Tht is hw u get the answer 945000J

In 2nd part this it would b neglected but in first it had to b considered!!!!

 

[littlecloud11]

Okay could u help me on this the curves. i drew them right till 2 sec mark. then i messed up each. any prediction how many marks ill loose?

for the v-t graph. 1 mark max. for d-t 2 likely, or 1 if you're lucky.

 

[littlecloud11]

In the part where velocity was CONSTANT, (i) u had to take the componemt of weight and add it to friction thn multiply by 400m!
Tht is hw u get the answer 945000J

In 2nd part this it would b neglected but in first it had to b considered!!!!

work done = mgh + WD against resistance = 1250*10* 400*.125 + 400* 800 = 945000J
i dnt see ne weight components here.

I guess the mgh serves as the weight component.

 

[Pals_1010]

work done = mgh + WD against resistance = 1250*10* 400*.125 + 400* 800 = 945000J
i dnt see ne weight components here.

I guess the mgh serves as the weight component.

How did you do the second part please?

 

[Gooners]

for the v-t graph. 1 mark max. for 2 likely, or 1 if you're lucky.

Yar please confirm that Q6(i) mai weight ka sin component along the plane downwards consider hona tha na !
Thts how u get 945kj!!!! Pls cnfirm!! ;p

 

[Saad Sarfraz]

for the v-t graph. 1 mark max. for 2 likely, or 1 if you're lucky.

Oh thxx..ur a genious!! btw what was ur distance in q3 the one in which we had to integrate!

 

[littlecloud11]

How did you do the second part please?

WD= change in GE + KE + WD against resistance
change is KE was 1/2* 1250* (10^2- 6^2)
you just add this to the previous ans.

 

[littlecloud11]

Yar please confirm that Q6(i) mai weight ka sin component along the plane downwards consider hona tha na !
Thts how u get 945kj!!!! Pls cnfirm!! ;p

as far as the formula goes,we dnt have to consider weight.

 

[littlecloud11]

Oh thxx..ur a genious!! btw what was ur distance in q3 the one in which we had to integrate!

Haha. No problem. It was 2.13m.

 

[Gooners]

How did you do the second part please?

As consant speed so force up the plane = force acting down the plane on object!!!!
800N + (12500Sin(7.18)) = 2362.33N
Now work = 2362.33 x 400 = 944934.5 = 945000J

Is mai 12500sin (7.18) is dwn the plane component!

 

[Saad Sarfraz]

Haha. No problem. It was 2.13m.

Lol. okay so mine was -2.13 somehow im seriously beginning to wonder was my M1 paper destined to be this bad. Plz pray i get around 35. P1 was good hoping to get around 69 inshAllah!!

 

[Gooners]

Haha. No problem. It was 2.13m.

as far as the formula goes,we dnt have to consider weight.

As consant speed so force up the plane = force acting down the plane on object!!!!
800N + (12500Sin(7.18)) = 2362.33N
Now work = 2362.33 x 400 = 944934.5 = 945000J

Is mai 12500sin (7.18) is dwn the plane component!

 

[Most_UniQue]

10.

Cool i gt same answer but you know in the last part were we calculate force i used sin a instead of replacing it with 0.125 and my answer using sin a is 984936 and if replaced with 0.125 the answer wud be 985000 so its a bit near to that , will i lose marks? My method and angle was right except that I used sin a instead of replacing it with 0.125

 

[littlecloud11]

Lol. okay so mine was -2.13 somehow im seriously beginning to wonder was my M1 paper destined to be this bad. Plz pray i get around 35. P1 was good hoping to get around 69 inshAllah!!

you got a negative after integration. Dnt worry, it probably wnt b as bad as u think.
I hope I can salvage an A* InshAllah.

 

[Pals_1010]

WD= change in GE + KE + WD against resistance
change is KE was 1/2* 1250* (10^2- 6^2)
you just add this to the previous ans.

Dang I forgot to add resistance to motion in the work done :/.... How many marks will I lose for that?

 

[littlecloud11]

As consant speed so force up the plane = force acting down the plane on object!!!!
800N + (12500Sin(7.18)) = 2362.33N
Now work = 2362.33 x 400 = 944934.5 = 945000J

Is mai 12500sin (7.18) is dwn the plane component!

the second part didn't hav constant speed!!!

 

[Most_UniQue]

As consant speed so force up the plane = force acting down the plane on object!!!!
800N + (12500Sin(7.18)) = 2362.33N
Now work = 2362.33 x 400 = 944934.5 = 945000J

Is mai 12500sin (7.18) is dwn the plane component!

I used weight component also so I gt 945000J and second answer was 985000

 

[Gooners]

you got a negative after integration. Dnt worry, it probably wnt b as bad as u think.
I hope I can salvage an A* InshAllah.

Hey yar!!!! Pls look at my method above and give it a seal of approval!!!!!!!

 

[littlecloud11]

As consant speed so force up the plane = force acting down the plane on object!!!!
800N + (12500Sin(7.18)) = 2362.33N
Now work = 2362.33 x 400 = 944934.5 = 945000J

Is mai 12500sin (7.18) is dwn the plane component!

So, bsically, we just solved the question using different methods.