Numerical Problem based on Newton's laws of Motion


by konichiwa2x
Tags: based, laws, motion, newton, numerical
konichiwa2x
konichiwa2x is offline
#1
Sep20-06, 07:16 PM
P: 81
A 5kg block is resting on the top right hand corner of a 10kg block. The length of the top of the 10kg block is 10m. Find the time taken by the 5kg block on top to slide of the 10kg block completely if the 10kg block is accelerating at 5m/s^2.

My work:

aB = accleration of the 5kg block
aA = acceleratipon of the 10kg block
If the 10kg block is accelerating to the right, then a pseudo force must be acting on the 5kg block on it in the opposite direction,

so fp = 5aB
aB = 5

aB - aA = -10

-10 = 1/2(-10)t^2
t = (2)^1/2

where am i going wrong? My book says 2 seconds is the correct answer. Please help
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Integral
Integral is offline
#2
Sep20-06, 07:24 PM
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P: 7,291
Since there is no friction in the problem, there is no force acting on the small block. Your "pseudo" acceleration should simply be the opposite direction of the acceleration of the large block. Your error is in adding the 2 accelerations. If you just use 5 [itex] \frac m {s^2} [/tex] as the acceleration the result is 2s.
konichiwa2x
konichiwa2x is offline
#3
Sep20-06, 07:30 PM
P: 81
ok so this pseudo acceleration is just the acceleration of the block B with respect to the block A, whereas the actual acceleration of the block B is 0. is that correct?

himanshu007
himanshu007 is offline
#4
May16-10, 03:10 PM
P: 1

Numerical Problem based on Newton's laws of Motion


since the lower block is of 10 kg and the upper block is just half its weight so the dimensions of the upper block will be half the lower block. so the length of the upper block is 5m.
now since there is no friction the lower block will have to move 10m so that tha upper block is completely fallen away.
now S = ut+1/2at^2
here u=0
a=5
s=10
so, 10=1/2 5t^2
t^2=4
t=2sec.


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