112)
Work Done by friction is W =- 50*10*0.5
W = -250 J
I think work will not be done by gravity and normal since the Displacement is perpendicular to them
1.. A bicycle is riding down a rough inclined plane with a uniform velocity with the rider applying brakes on only the rear wheels. Giv the direction of friction on both the front and the back wheels.
2.. A man of mass 50kg starts walkin from rest with and attains a velocity of 10 m/s after coverin a distance of 20. coef of friction is .5. find the work done by friction, normal reaction and gravity.
3.. A conical pendulum setup is given. About the point of suspension which of the following are correct...?
a.. Magnitude of Angular momentum remains constant
b.. Direction of angular momentum remains constant
c... Angular momentum remains constant .
d... Torque abt point of suspension is zero.
4.. A particles of mass 1kg slides ion a circle 20m with a speed 1m/s. Only forces in vertical are Weight and normal reaction. coef of friction =.5 No data is available about the forces in horizontal. Find the direction of friction and the magnitude .
Plz dont be in hurry to post ur ans these are not as easy theu luk to be.... but plz do post them
-
UP 0 DOWN 0 2 21
21 Answers
11
1yaar friction tab lagagi when foot is on the ground....and the foot doesn't move in that condition along the ground....thus work done by the friction will also be zero.....
11Q2
THE WORK DONE BY THE FRICTION WILL BE PRESENT BUT THE WORK DONE BY NORMAL AND THE GRAVITY WILL BE 0 AS THEY R PERPENDICULAR TO THE DISPLACEMENT
WORK DONE BY FRICTION =μN
=0.5 X 50 X 10
=250J
62I think what mani has done is correct
I will check it again though
11BHAI CONDITION AFTER BRAKING
MY JUSTIFICATION : AS WE APPLY BRAKES THE REAR TYRE WILL SUDDENLY STOP TO MOVE
ITS ANTICLOCKWISE TORQUES WILL TEND TO STOP BUT THE TYRE CONTINUE TO MOVE FORWARD
THAT IS THE REASON WHY I HAVE SHOWN FRICTION BACKWARD
BUT AS THE FORWARD TYRE CONTINUE TO MOVE FORWARD SO FRICTION ON IT MUST REMAIN UNCHANGED
11@DEEPANSHU SAID IN #11
ok ... i appreciate ur effort
but the ans given is that there is no friction on front wheel and friction on rear while wheel is
upwards as correctly said by u ...
can u explain ur ans when he s not braking .... i think it should opposite to wat u ve shown on both tyres
LETS TALK ABOUT THE REAR WHEEL BEFORE THE APPLICATION OF BRAKES
CONCEPT : FRICTION FORCE DEPENDS UPON THE POINT OF CONTACT RATHER THAN THE MOTION OF THE WHOLE BODY
LOOK AT THE REAR WHEEL AND THE BICYCLE IS GOING DOWNWARDS
DUE THE MOTION OF THE BICYCLE AN ANTICLOCKWISE TORQUE COMES INTO PLAY WHICH TRIES TO ROTATE THE WHEEL BACKWARDS BUT AS FRICTION IS PRESENT IT WILL OPPOSE ITS MOTION AND WILL ACT IN THE FORWARD DIRECTION
NOW A QUESTION ARISES IF THERE IS NO FRICTION WHAT WILL BE THE MOTION OF THE CYCLE ?
THE ANSWER IS BICYCLE WILL MOVE BACKWARDS
BUT I CANT UNDERSTAND UR ANSWER THAT WHY THERE WILL BE NO FRICTION ON THE FRONT WHEEL AFTER THE APPLICATION OF BRAKES??
IF I AM WRONG PLEASE CORRECT ME EXPERTS
13ok ... i appreciate ur effort
but the ans given is that there is no friction on front wheel and friction on rear while wheel is
upwards as correctly said by u ...
can u explain ur ans when he s not braking .... i think it should opposite to wat u ve shown on both tyres
11DEEP....SAID
@mani
i m not getting wat do u mean by after and before...
DUDE I AM TAKING THE CONDITIONS BEFORE BRAKING AND AFTER BRAKING
13Found them in fiitjee rtpf i dont agree with sum of the anss given...
13@mani
i m not getting wat do u mean by after and before...
@sky
ans to the third ques is only a option (as given) . i think v r wrongly interpretin the words
'abt point of suspension'
for 2nd ques ans given is that work by ol the 3 forces is 0....
i dont understandd how..
sum1 try the lasy ques plz..
13sry see the tym i posted the ques
i corrected the options nao....
11DEEPANSHU SAID.............
3.. A conical pendulum setup is given. About the point of suspension which of the following are correct...?
a.. Angular momentum remains constant
b.. Direction of angular momentum remains constant
c... Angular momentum remains constant .
d... Torque abt point of suspension is zero.
MY DEAR WHAT IS THE DIFFERENCE BETWEEN A AND C [7][7][7]
12.) there will be work done by friction.. baki doh ka zero hoga [waise.. i think i am not wrong!]
13) Magnitude of angular momentum about the point of suspension remains constant.
But is it (a) or (c)[7]