Physics help ???? 2 Questions constant acceleration?

Question

Could someone help me with these problems, because I have a test coming up and these are the suggested problems. Thank you

1) An airplane lands and starts down the runway at a southwest velocity of 45 m/s. What constant acceleration allows it to come to a stop in 1.4 km

2) Two blocks are connected by a lightweight, flexible cord that passes over a frictionless pulley. If m1 is 3.4 kg and m2 is 9.1 kg, and block 2 is initially at rest 140 cm above the floor, how long does it take block 2 to reach the floor?

Answer 1

I should be able to help as I am a mechanical engineering student and work with this kinda stuff :)

tip:ALWAYS draw diagrams and pictures to help you find out whats going on.

1)You can see the acceleration will be negative as it has
to slow the plane down(positive would speed the plane up)

use your newtons equations. dont worry bout the direction.
(vxv)=(uxu) +2as
v=0(final velocity)
u =45(intial velocity)
s=1.4km (distance) convert to 1400m
a= constant acc

therefore a= [(vxv)-(uxu)] / 2s
= [(0)-(45x45)] / (2x1400)
= -0.723 m/sxs (constant acc)

2)once again DRAW pictures of the question
assuming the cord is inflexible
Use newtons equations again
F=mxa
F=force, m= mass, a= acceleration

for mass 1
F=3.41 x 9.81 (a=gravity=9.81)
= 33.4521

for mass 2
F = 9.1 x 9.81
= 89.271

As both masses are moving in opposite directions, subtract their forces to find the total force and its direction.
F=89.27-33.45 = 55.821N (this is the total force acting on mass 2 now)

now we can find the acceleration acting on m2.
F=ma
a= F/m
= 55.821/9.1
= 6.134m/sxs

we know from newton that (vxv)=(uxu) + 2as
therefore (vxv)= 0 + 2(6.134)(1.4cm)
v= 10.26m/s
This v is the velocity as the mass hits the floor
Now we use newtons other equation
v=u + at
v=10.26(final velocity is when as m2 hits floor)
u=0(initial)
a=6.134m/sxs
t= required time

therefore t=(10.26-0)/6.134
= 1.67 s (time taken for block to hit floor)

Any ways that should be correct. I may have made some calculator errors, but the method is important. you get marks for method even if the answer is wrong in your tests. thats how i passed science :)

Apoligies if anyone noticed some really stupid mistakes. i get impatient

Answer 2

Under constant acceleration a, the distance traveled is .5*a*t^2. The veloctiy is v = v0 - a*t. When the airplane stops, v = 0, so a*t = v0. Put this in the first equation s = .5*v*t. Therefore the time it takes to stop is t = 2*s/v0. Put this into the second equation to get v0 = a*2*s/v0; solve for a = (v0^2)/(2*s)

The force on m2 is its weight minus the weight of m1, or (m2-m1)g; the acceleration is a = f/m2=(m2-m1)g/m2. Again using the ditance/acceleration formua s=.5at^2, t = sqrt(2s/a); put s (140cm) and a = (m2-m1)g/m2 into this formula to get t

Answer 3

heehee I am taking physics ^^

1) well for this question just use the formula:
(Vf)^2 = (Vi)^2 + 2*acceleration*distance
2) flexible cord? huh? you mean non-flexible right?
if non-flexible then...
F = 9.1*acceleration of the system
F = 9.1*gravity - 3.4*acceleration of the system
Solve for "a" with the two equations. Then find the time with this equation:
Distance = Vi*time + 0.5 * a * (t)^2
Note that in the problem initial velocity is zero.

Not sure if it's right, but I think it is.

Answer 4

'passes via' implies a precise preliminary pace, which you have not discovered within the predicament. The function x = xi + Vi*t + ½*a*t² = zero + Vi*three + ½(-two)*three² = 3Vi - nine wherein Vi < zero