Friction and Forces. (mcq questions)

[s200960170]

Qs1)
A constant horizontal force of 36 N is acting on a block of mass 4.0 kg, another block of
mass 2.0 kg sits on the 4.0 kg block. The 4.0 kg block moves on a frictionless horizontal
floor. Find the magnitude of the frictional force maintaining the 2.0 kg block in its position
above the 4.0 kg block during the motion.
A) 12 N
B) 15 N
C) 18 N
D) 16 N
E) 36 N


Qs2)
Two horizontal forces of equal magnitudes are acting 011 a box sliding on a smooth horizontal table. The direction of one force is the north direction; the other is in the west direction. What is the direction of the acceleration of the box?
A) 45° west of north
B) 20° east of north
C) 60° north of east
D) 45° south of east
E) 10° west of south

[how on Earth do you even approach these problems, please help]

 

[tiny-tim]

welcome to pf!

hi s200960170! welcome to pf!

for 1), do a https://www.physicsforums.com/library.php?do=view_item&itemid=100" for each block, showing all the forces on that block

for 2) tell us what you think, and then we'll comment

 

[tiny-tim]

A stone is thrown vertically up from the edge of the top of a 100-m high building. It reaches the ground (at the bottom of the building) after 10.0 s. What is the initial speed of the stone?

A) 39.0 m/s
B) 29.0 m/s
C) 49.0 m/s
D) 59.0 m/s
E) 69.0 m/s

[PLEASE HELP]

use one of the standard https://www.physicsforums.com/library.php?do=view_item&itemid=204" equations …

what do you get? ​

 

[s200960170]

I figured out the stone problem, thank you very much Tiny-tim :)

t = 10s
xf-xi = -100m
g= -9.8

so when we use the formula x=vi*t + 0.5at^2...
i got 39m/s :)

 

[tiny-tim]

yup!

(are you ok on the other two questions?)

 

[s200960170]

Qs1, i understood it. i used the formula a = f/m = 36 / 6 = 6 m/s/s so frictional force on the 2 kg must be f=ma. => 2*6 = 12 N which is choice A)

for Qs, 2 after drawing the diagram and drawing arrows representing the forces, the picture became very clear. because the forces have equal magnitude, the angle has to be 45 degree in the direction WN or NW.

thank you. for your help.

there is however one more Qs, I am finding it difficult to approach, Qs is
- At what angle should the circular roadway of 50m radius, be banked to allow cars to round the curve without slipping at 12m/s (friction is ignored)?

 

[tiny-tim]

- At what angle should the circular roadway of 50m radius, be banked to allow cars to round the curve without slipping at 12m/s (friction is ignored)?

again you need to draw a https://www.physicsforums.com/library.php?do=view_item&itemid=100" …

this time there will be only two forces, but they will have to equal mass times the https://www.physicsforums.com/library.php?do=view_item&itemid=27"

 

[s200960170]

im sorry i didnt quiet follow,
the mass of the object is not given only the velocity and radius is given.
can you please tell me as to which formula to use?
is it F = m(v^2/r)

 

[tiny-tim]

call the mass "m" … it wil cancel out in the end

(and yes, v²/r for the acceleration)

 

[s200960170]

your torturing me giving me small pieces of cake, while i would like to have the whole dish at once, XD
i'm still unable to figure out the solution,

1st you said there will be to forces,
so i assumed the forces are ma, and mg(sin(teeta))
and i used Newtons second law, but its not working :(..
please help and this time please try to be a bit more descriptive. :)
thank you :P

 

[tiny-tim]

have you drawn a free body diagram?

the two forces are the weight mg, and the normal https://www.physicsforums.com/library.php?do=view_item&itemid=73" R …

together they have to equal mv²/r horizontally

 

[s200960170]

mg(cos(teeta)) is the normal force,
and YES i drew the diagram, and i don't see how "mg"and the normal force can be connected by a formula.

 

[s200960170]

Hey Tim, If the magnitude of a vector is given how do i find its components?

 

[tiny-tim]

(have a theta: )

mg(cos(teeta)) is the normal force,

uhh? why?

 

[s200960170]

- At what angle should the circular roadway of 50m radius, be banked to allow cars to round the curve without slipping at 12m/s (friction is ignored)?

i found the formula i was looking for :tongue2:
and now i don't have to draw any forces or anything just substitute the values and you get the answer,
the formula is --> "v^2 = rg tan(theeta) " its that's simple,

However, now I am stuck in a vector problem,

Consider two vectors A and B with magnitude 5cm and 8cm, respectively. Vector A is along the positive x-axis and vector B is along the positive y-axis. Find A*(A+B).

[where the symbol * stands for dot product].

Any idea How to tackle that problem?

 

[tiny-tim]

i found the formula i was looking for :tongue2:
and now i don't have to draw any forces or anything just substitute the values and you get the answer,
the formula is --> "v^2 = rg tan(theeta) " its that's simple,

hold it!

you'll never remember that formula for the exam,

and anyway you do need to be able to do this sort of problem …

before you leave this problem, try to work out why the normal force is different in this case

 

[s200960170]

My major exam for Physics is tomorrow I am sure i can have that formula in my mind till then lol, but i agree your right its not the right thing to do, to mug up formulas without understanding the concept :)

so i know why the normal force is different,
its because the normal force is the force perpendicular to the ground, and since the road is banked the car's nomal force is not mg, that's why i resolved the downward force ie "mg" into components , "mg cos(@)" and "mg sin(@)".

regarding the vector problem, its funny how after reading the problem after posting it on this forum i was able to figure it out, the key idea is the the words "along the positive x-axis" and "along the along the positive y-axis" :P

 

[tiny-tim]

hi s200960170!

actually it's all to do where the zero acceleration is …

on a flat slope, the acceleration normal to the plane has to be zero, so the normal force equals mgcosθ

but on a curved slope, the centripetal acceleration is horizontal, so it does have a normal component, so that doesn't work: in this particular case, we know that the vertical acceleration is zero, so mg equals the normal force times cosθ

good luck in your exam tomorrow ​