Physics, asked by debajitsen160, 8 months ago

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​

Answers

Answered by Anonymous
47

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

Answered by Anonymous
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}

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