A block of metal 4 kg is in rest on a frictionless surface. It was targeted by a jet releasing water of $2…
- $10 \mathrm{~ms}^{-2}$
- $15 \mathrm{~ms}^{-2}$
- $20 \mathrm{~ms}^{-2}$
- $5 \mathrm{~ms}^{-2}$
Solution

$\mathrm{v}=10 \mathrm{~m} / \mathrm{s}, \frac{\mathrm{dm}}{\mathrm{dt}}=2 \mathrm{~kg} / \mathrm{s}$
$\therefore \quad$ Force applied by jet on the block is $\mathrm{F}=\mathrm{v} \cdot\left(\frac{\mathrm{dm}}{\mathrm{dt}}\right)=10 \times 2=20 \mathrm{~N}$
$\therefore \quad$ Acceleration of the block is
$\mathrm{a}=\frac{\mathrm{F}}{\mathrm{~m}}=\frac{20}{4}=5 \mathrm{~m} / \mathrm{s}^2$
Asked in: AP EAMCET 2024 (21 May Shift 2)