A Couple of Mechanics Questions

[MAVtKnmJ]

Q) A particle P is held at rest at a fixed point O and then released. P falls freely under gravity until it reaches the point A which is 1.25m below O.

(i) Find the speed of P at A and the time taken for P to reach A.

The particle continues to fall, but now its downward acceleration t seconds after passing through A is (10 − 0.3t)ms^-2

(ii) Find the total distance P has fallen, 3 s after being released from O.

I've solved part one to get...

t=0.5s and v=5ms^-1

But I'm unable to do part (ii). Don't know where to start and how to solve this. I've tried to integrate dv/dt but not getting anywhere.

Thanks!

 

[MAVtKnmJ]

Q2) A lorry of mass 15 000 kg moves with constant speed 14ms−1 from the top to the bottom of a straight hill of length 900 m. The top of the hill is 18m above the level of the bottom of the hill. The total work done by the resistive forces acting on the lorry, including the braking force, is 4.8 × 106 J.

Find

(i) the loss in gravitational potential energy of the lorry
(ii) the work done by the driving force.


My Attempt

The answer to part (i) is mgh = 2700000J

But I am stuck at (ii)

I used the formula

WDF - Work Done By Resistance = 2700000

WDF = 2700000+4.8x10^6
WDF = 7500000J

But the correct answer is 2.1 x 10^6J

How? Am I using a wrong method?

 

[Zazzyo]

To the above question ^^^

Since the speed is constant,

Workdone by driving force = workdone gainst resistance + Chancge in P.E
WDF = R + (-E.P) << bcuz it is loss it will be -ve

I tried to use conservation of energy.. however i do not use specific formlae for doing m1 questions..

 

[MAVtKnmJ]

Great, thanks for the quick reply.

What about the first one?

 

[1992]

ok ive done the question, is the answer 31.7m??? to make sure ive done it right???

im going to explain the procedure:
First integrate the aceleration to find speed: V= 10t -015t^2
Then integrate again to find distance: s= 5t^2 - 0.15t^3/3 +c , use the values t=0.5 and S=1.25, substitute inside this equation. and u get c= 6.25* 10^-23

Now u need to find the distance travelled in 2.5 Seconds only, becuase as u calculated in part i) it took 0.5sec to reach that point right-?' ( 3-0.5=2.5)
so substitute 2.5 in here: 5t^2 -0.05t^3 + 6.25*10^-23
Ans gives u 30.5M. total distance is 1.25+30.5= 31.7m

 

[allah sareej]

can anyonen help me on
may/june 2005
q2
really stck on ds
thnks in advance

 

[spanishswan]

june 05 q2

its like first u have to find the forces in the y direction and x direction [axis]
for y its 5sin50 - 6sin 30 = 0.8302
for x its 7 + 5cos50 - 6cos30 = 5.018
now resultant is under root of x square + y square
so its 5.09
n the direction is the angle...which is tan theta = y/x
which will give u 9.4 degrees ..which we take as anticlockwise from x axes.

 

[spanishswan]

for the very first question asked on the page...the answer is 44.2 not 31.7
u went rong at findind the velocity equation...cz theres a constant c as well...
so the equation will be 10t - 0.5t^2 + c
at t=0, velocity is 5..from part 1..the particle reached A at this velocity
so putting t as zero n v as 5 u l get the value of c which is 5
now proceed as said..integrate to find distance with limits 0 and 2.5...n add 1.25 for the distance covered till A..u l get the answer
does this help??

 

[blackswan7]

for the very first question asked on the page...the answer is 44.2 not 31.7
u went rong at findind the velocity equation...cz theres a constant c as well...
so the equation will be 10t - 0.5t^2 + c
at t=0, velocity is 5..from part 1..the particle reached A at this velocity
so putting t as zero n v as 5 u l get the value of c which is 5
now proceed as said..integrate to find distance with limits 0 and 2.5...n add 1.25 for the distance covered till A..u l get the answer
does this help??

why do you use t=0, shouldn't it be t=0.5?