Question

two blocks X and Y of masses m and 2m respectively are accelerated along smooth horizontal surface by a force applied to block X.what is the magnitude of force exerted by block y on block X during this acceleration

Answer 1

Answer:

2F/3.

Explanation:

Both blocks will be accelerated when the block X is applied by a force of magnitude F. Since, the total mass will be m+2m which is 3m.

So, the acceleration of the system will be given as F=ma or a=F/3m. So, from this we can find the force acting on the block Y. So, to calculate the force on the block Y we have that F(y) =2m*a or 2m*F/3m which on solving we will get the force on the block y as F(y)= 2F/3.

Answer 2

Answer:

2F/3

Explanation:

GIVEN DATA:

mass of x=m

mass of y=2m

TO FIND:

force exerted by Y=?

SOLUTION:

the total mass is m+2m=3m

now f=ma

a=F/m  

as  m=3m so put

a=F/3m

now f(Y)=ma

as mass of m(y) is 2m & a=F/3m so put:

F(Y) =2m*F/3m

F(Y)=2F/3