Question

a plank of mass 10 kg rests on smooth horizontal surface two Block A and B of mass 2 kg and 1 kg lies at a distance of 3 metre​

Answer 1

A plank of mass 10 kg rests on a smooth horizontal surface 2 blocks a and b of masses 2kg and 1 kg respectively lies at a distance of 3 metre on the plank the friction coefficient between the blocks and planks are equal to 0.3 for block a and equal to 0.1 for block b. now a force f is equal to 50 newton is applied to the plank in the horizontal direction . find time in seconds after which block a collides with block b.

Answer 2

Can't be!

Look at this:

for a, the limiting friction = (0.3)(2kg)(10m/s2) = 6N,

so, any external force above 6N FOR the block should make it move, which means to say, a force must exceed 39N to make the block moove.

Now, for 50N, both the blocks will move. This will make us to calculate a relative acceleration. And so we got 1.73 seconds.

But, for 6 seconds, the external force must be close to 15.17 and not 50 newtons... lemmie tell u why...

Suppose, block a is at rest:

so, mA = net force(A is acceleration on the plank) = ma - k.mg

m- mass of small plank

a = Fnet/mnet =F/(13kg) = acc. of small block when the plank with the two blocks is pushed

k = 0.1

also, A = 2l/t2 (l-distance of sep., t =6 seconds... we pressume)

=(2)(3)/(6)(6) = 1/6 m/s2

so, mA = ma - kmg

A = a -kg

a = A +k.g

F/13 = 1/6 +(0.1)(10)

=7/6

so, F = 91/6 ~ 15.17N and not 50N!

So, I suggest, you recheck your question... then, see if the answer, if it is still 6 seconds... we both would be waiting for an expert to come by...

yepyepyep!!! I really wanna know how to solve this question... maybe... something more complicated must be used...