friction between the two blocks:
Ff=60*3.7
well, this Ff drives the lower block.
Ff=ma=100*a or
a=60*3.7/100=2.22 acceleaetion of lower block.
so the relative acceleration on the top block=3.7-2.22=1.5m/s^2
to time the blocks are in contact
d=1/2 a t^2 or
t= sqrt(2d/a)=sqrt(2*...hm, wondering what the distance is along the lower block?
A 60 kg block slides along the top of a 100 kg block. The lighter block has an acceleration of 3.7 m/s2 when a horizontal force F= 350 N is applied. Assuming there is no friction between the bottom 100 kg block and the horizontal frictionless surface but there is friction between the blocks. Find the acceleration of the 100 kg block during the time the 60 kg block remains in contact.
1 answer
1 answer