acceleration of the block


A force F acts on a block of mass m placed on a horizontal smooth surface at an angle Θ with horizontal. Then

(A) If F sinΘ < mg then a = F + mgm
(B) Acceleration = FCosΘm when F > mg cosecΘ
(C) Acceleration = FM if F SinΘ > mg
(D) If F sinΘ > mg then a = F + mgm

10 Answers

36
rahul ·

Hey!! the image i uploaded is not visible.... what's wrong with this site...
Admin... u must bring some improvement in this site.. :{

71
Vivek @ Born this Way ·

I have seen similar problem in Irodov also. And Yes the images are not working. I had to use tinypic.

36
rahul ·

thanks to tinypic :p

1
Nishant Consul ·

What to do and what not to do in the last month before IIT-JEE

http://www.youtube.com/watch?v=Z5-AfiJxjOc&context=C4acfebbADvjVQa1PpcFMQX-FKsRHXxJDA7RTt_k4sN1NL6Iapsuo=

1
funkygp ·

is it d?

36
rahul ·

ya d ....

butw i m getting

a = (F - mg) / m

71
Vivek @ Born this Way ·

It's because, though you have used the vectors, you have used the minus sign in front of mg. If you are using vectors, you can always add them without thinking too much for the sign.

36
rahul ·

"though you have used the vectors, you have used the minus sign in front of mg"

ok

"If you are using vectors, you can always add them without thinking too much for the sign."

didn't get this?

please explain...!!

do u mean that it won't be bringing any change in magnitude.. or something else..?

as in both the cases the magnitude has to be √(F)2 + (mg)2m

262
Aditya Bhutra ·

@rahul - mg is a vector which is pointing vertically downward, so why are you applying the negative sign before it ??

36
rahul ·

on drawing the F.B.D we get,

F sinΘ + N - mg = may

Now, if F sinΘ > mg then N vanishes

so, F sinΘ- mg = may --- (i)

Again, F cosΘ = max ---- (ii)

Adding these two we get,

F sinΘ + F cosΘ - mg = may + max

or, ax + ay = (F - mg) / m

that's the reason i guess....!!

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