Needs help to some Mechanics questions

[Amy Bloom]

I would be very grateful if anybody could help me in solving those questions please.

1) A particle starts from rest and slides with acceleration 3ms-2 down a fixed plane which is inclined at 40 degrees to the horizontal. Calculate, to 2 significant figures the coefficient of friction between the particle and the plane. (i can't get the answer!!! ) (0.45)

I can't find the last part of this question...
2) A balloon rises vertically from rest on the ground with constant acceleration 1.25 ms-2. A small stone is released from the balloon when it has risen to a height of 250m. find:
(a) The speed of the balloon when the stone is released (25 ms-1)
(b) The time taken by the stone to reach the ground (10s)
(c) The greatest height of the stone above the ground (281 m)
(d) The speed of the stone on reaching the ground (75 ms-1)
(e) The height of the balloon above the ground at the moment the stone hits the ground (563 m)

 

[darknessinme]

For q1:
(Weight down slope)-(friction up slope)= resultant force
(mg sin 40)-(u x R)=3xm
(mg sin 40)-(u x mg cos 40)=3m
Simplify to get: g(sin 40 - cos 40 u)= 3
Solve for "mew" so u=0.447....=0.45 2sf

 

[Amy Bloom]

For q1:
(Weight down slope)-(friction up slope)= resultant force
(mg sin 40)-(u x R)=3xm
(mg sin 40)-(u x mg cos 40)=3m
Simplify to get: g(sin 40 - cos 40 u)= 3
Solve for "mew" so u=0.447....=0.45 2sf

Thank you soo much!

 

[Amy Bloom]

Can somebody solve this for me:
A block of mass 3 kg is pulled along a rough horizontal floor by a constant force of magnitude 20N inclined at an angle of 60 degrees to the upward vertical (explain to me what this means ). The acceleration of the block has magnitude 2 ms-2. Calculate to 2 decimal places, the value of the coefficient of friction between the block and the floor. [Answer: 0.57]

 

[SmartNour]

60 degrees to the vertical means you gotta inverse the formulas to find the horizonatal and vertical component in this case Horizontal component fx = fsin (angle) and the vertical component fy=fcos(angle).
to find the coefficient of friction MU=F/R
so you have to find the friction first and find R( the normal contact force )

 

[SmartNour]

btw what do we

Can somebody solve this for me:
A block of mass 3 kg is pulled along a rough horizontal floor by a constant force of magnitude 20N inclined at an angle of 60 degrees to the upward vertical (explain to me what this means ). The acceleration of the block has magnitude 2 ms-2. Calculate to 2 decimal places, the value of the coefficient of friction between the block and the floor. [Answer: 0.57]

have to fiind the coeffiecient of friction is already given

 

[Amy Bloom]

btw what do we
have to fiind the coeffiecient of friction is already given

That's what i did! Can u please post a drawing or smething like that for me to have a clear picture of it?
Oh U mean the 0.57?! Its the answer to the question, for ppl to know when they tried it they got the right answer.
Thanks a lot!

 

[SmartNour]

That's what i did! Can u please post a drawing or smething like that for me to have a clear picture of it?
Oh U mean the 0.57?! Its the answer to the question, for ppl to know when they tried it they got the right answer.
Thanks a lot!

ehm ehm ok fine i will try posting asap

 

[Amy Bloom]

ehm ehm ok fine i will try posting asap

Thanx!

 

[anzaar]

A block of mass 3 kg is pulled along a rough horizontal floor by a constant force of magnitude 20N inclined at an angle of 60 degrees to the upward vertical (explain to me what this means: ). The acceleration of the block has magnitude 2 ms-2. Calculate to 2 decimal places, the value of the coefficient of friction between the block and the floor.
solution:
this Means force makes 60 degree with the vertical axis. we know there is 90 degree angle between horizontal and vertical axis so the force makes 40 degree angle with the x- axis.
here is the component of constant force along x axis: 20 cos40 and vertical component of force, which is R=20sin40
let frictional force is f, so net force is
20 cos40-f=ma
=3*2
evaluate" f" from here . then use formula f=coefficient of friction x normal reaction R , where R=20sin40 ( means vertical component of force)
hence you will get the answer.

 

[Amy Bloom]

A block of mass 3 kg is pulled along a rough horizontal floor by a constant force of magnitude 20N inclined at an angle of 60 degrees to the upward vertical (explain to me what this means: ). The acceleration of the block has magnitude 2 ms-2. Calculate to 2 decimal places, the value of the coefficient of friction between the block and the floor.
solution:
this Means force makes 60 degree with the vertical axis. we know there is 90 degree angle between horizontal and vertical axis so the force makes 40 degree angle with the x- axis.
here is the component of constant force along x axis: 20 cos40 and vertical component of force, which is R=20sin40
let frictional force is f, so net force is
20 cos40-f=ma
=3*2
evaluate" f" from here . then use formula f=coefficient of friction x normal reaction R , where R=20sin40 ( means vertical component of force)
hence you will get the answer.

Hey thanks!

 

[anzaar]

welcome

 

[Amy Bloom]

Hey there! Two questions are really annoying me. I understood the questions but I'm not getting the answers . Can anyone gimme a helping hand please?

 

[VelaneDeBeaute]

Why don't you post the method how you did the question, and let others find the flaw?
P.S. I'm getting a = 1 m/s^2 (quite a rough estimate), and T = 18N for the first question!

 

[Amy Bloom]

Why don't you post the method how you did the question, and let others find the flaw?
P.S. I'm getting a = 1 m/s^2 (quite a rough estimate), and T = 18N for the first question!

I shall do that. I'm sorry its not the answer. I'll post my workings and u check. Gimme a few minutes. So here u are:

 

[Amy Bloom]

Hey there, can anybody help me with this please? I would be very grateful.
Help me with question 5 and check if my workings for question 7 is correct please.

 

[anzaar]

Question 5
(i) A cyclist travels in a straight line from A to B with constant acceleration 0.06 ms-2 . His speed at A is 3 ms-1 and his speed at B is 6 ms-1. Find:
(a) the time taken by the cyclist to travel from A to B
(b) the distance AB
Ans:
(a)
v=u+at
6=3+.06t
t=50s
(b) 2as=v2 -u2
2 x 0.06s=36-9
s=27/0.12 m

 

[Amy Bloom]

Question 5
(i) A cyclist travels in a straight line from A to B with constant acceleration 0.06 ms-2 . His speed at A is 3 ms-1 and his speed at B is 6 ms-1. Find:
(a) the time taken by the cyclist to travel from A to B
(b) the distance AB
Ans:
(a)
v=u+at
6=3+.06t
t=50s
(b) 2as=v2 -u2
2 x 0.06s=36-9
s=27/0.12 m

Hey thanks ^^, question 7 is correct?

 

[anzaar]

welcome

 

[Amy Bloom]

welcome

Hey anzaar, could you solve the (ii) of question 5? Just need a helping hand.
According to you, what i did for question 7 is it correct?