Question

Find acceleration?

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.

Answer 1

`a = 1.3` `"m/s"^2`

Explanation:

A force of `350` `"N"` is applied to two objects with different masses and therefore produce different accelerations.

If the block of mass `60` `"kg"` has an acceleration of `3.7` `"m/s"^2`, then the net force acting on it is

`SigmaF = ma = (60color(white)(l)"kg")(3.7color(white)(l)"m/s"^2) = color(red)(222` `color(red)("N"`

The remaining part of the `350`-`"N"` applied force `F` is in the friction force `f` between the two blocks:

`f = F-F_ "60 kg" = 350` `"N"` `- color(red)(222` `color(red)("N"` `= color(green)(128` `color(green)("N"`

This is the net force that causes the acceleration of the `100`-`"kg"` block. The acceleration of this block is thus

`a = (SigmaF)/m = (color(green)(128)color(white)(l)color(green)("N"))/(100color(white)(l)"kg") = color(blue)(1.3` `color(blue)("m/s"^2`

which I'l leave rounded to `2` sig figs.