Figure shows a block of mass `m_(1)` sliding on a block of mass `m_(2)`, with `m_(1) gt m_(2)`. Find the

(a) acceleration of each block
(b) tension in the string
(c ) force exerted by `m_(1)` on `m_(2)`. ltbr. (d) force exerted by `m_(2)` on the incline.

Answer

(a) `a=(m_(1)gsintheta-m_(2) g sin theta)/(m_(1)+m_(2))` (b) `(2m_(1)m_(2)g sin theta)/(m_(1)+m_(2))` (c ) `m_(1)g cos theta` (d) ` (m_(1)g cos theta+m_(2)g cos theta)`

Explanation

Figure shows free body diagram of each block. We will apply Newton's second law along x- and y-axis shown in free body diagram. Block `m_(1)` is heavy, hence it slides down whereas `m_(2)` slides up.

`Sigma F_(x)=m_(1)g sin theta-T=m_(1)a`
`Sigma F_(y)=N_(1)-m_(1)g cos theta=0`
`Sigma F_(x)=T-m_(2)g sin theta=m_(2)a`
`Sigma F_(y)=N_(2)-N_(1)-m_(2)g cos theta=0`
From eq. (i) and (iii),
`a=(m_(1)g sin theta-m_(2) g sin theta)/(m_(1)+m_(2))`
`T=m_(2)a+m_(2)g sin theta`
`=(m_(2)(m_(1)g sin theta-m_(2)sin theta))/(m_(1)+m_(2))+m_(2)g sin theta`
`=(2m_(1)m_(2)g sin theta)/(m_(1)+m_(2))`
From Eqs. (ii) and (iv)
`N_(1)=m_(1)g cos theta`
`N_(2)=N_(1)+m_(2)g cos theta = (m_(1)g cos theta-m_(2)g cos theta)`
`N_(1)` is force exerted by block `m_(1) on m_(2) and N_(2)` is force exerted by block `m_(2)` on inclined surface.