Question

calculate the value of force that is applied for 1 second on a 10kg object to increase it's velocity from 5m/s to 10m/s
plz help me​

Answer 1

Answer:

$\boxed{\sf Force \ applied \ on \ object = 50 \ N}$

Given:

Mass of object (m) = 10 kg

Initial velocity (u) = 5 m/s

Final velocity (v) = 10 m/s

Force applied for time interval (t) = 1 sec

Explanation:

As, we know;

$\sf \implies Force = \frac{\Delta p}{\Delta t} \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{\Delta (mv)}{\Delta t} \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{m\Delta v + v \Delta m}{\Delta t}$

$\sf As \ mass \ of \ the \ object \ is \ constant; \\ \sf Therefore, v \Delta m = 0 \\ \\ \sf \implies Force = \frac{m\Delta v}{\Delta t} \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{m(v - u)}{ \Delta t} \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{10(10 - 5)}{1} \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 10 \times 5 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 50 \: N$

Answer 2

Given ,

Time (t) = 1 sec

Mass (m) = 10 kg

Initial velocity (u) = 5 m/s

Final velocity (v) = 10 m/s

We know that ,

$\star \: \: \sf F = ma \\ \\\star \: \: \sf F = m( \frac{v - u}{t} )$

Thus ,

$\sf \mapsto F = 10( \frac{10 - 5}{1} ) \\ \\\sf \mapsto F = 10 \times 5 \\ \\\sf \mapsto F = 50 \: \: newton$

$\therefore \sf \underline{The \: force \: applied \: by \: an \: object \: is \: 50 \: N}$