91Ó°ÊÓ

A steel cable lying flat on the floor drags a $20kg$block across a horizontal, frictionless floor. A $100N$force applied to the cable causes the block to reach a speed of $4m/s$in a distance of $2m.$

Force of block is obtained by : $F_{\text{block}}=m_{1}a⋯-\left(1\right)$

Now, to find a, by kinematic equations we get,

$$\begin{gathered} v^2-u^2=2aS \\ ⇒a⇒=\frac{v^2-u^2}{2S}⋯\left(2\right) \end{gathered}$$

Now equation (1) localid="1650272521830" $F_{\text{block}}=m_1\frac{v^2-u^2}{2S}⋯\left(3\right)$

The same force will apply on cable in backward direction . So net force on cable can be expressed as follows.

localid="1650272827260" $⇒\text{Resultant force}{F}_{\text{cable}}=F_{\text{ext}}-F_{\text{block}}⋯\left(4\right)$

Substituting the given data in equation (1) and (2) we get,

localid="1650272864743" $$\begin{gathered} ⇒a=\frac{4^2-0^2}{2×2} \\ ⇒a=4\mathrm{m}/{\mathrm{s}}^2 \\ {F}_{\text{block}}=20kg×\frac{\left|{(4m/s^2)}^2-0^2\right|}{2m×2m} \\ ⇒F_{\text{biock}}=80\mathrm{N}⋯\left(5\right) \end{gathered}$$

Substituting the data in equation (5) we get,

localid="1650272911920" $$\begin{gathered} ⇒F_{cable}=100-80 \\ ⇒F_{cabie}=m_{2}a=20\mathrm{N} \\ ⇒m_2=\frac{20}{4}=5\mathrm{kg} \end{gathered}$$

Mass of the cable is$5kg.$