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`        A constant force acts on a object of mass 2kg for 10sec and increases its velocity from 5m/s to 10m/s. Find the magnitude of applied force . If this force was applied for a duration of 15sec, what would be the velocity of the object`
one year ago

```							Change in velocity=10-5=5m/sTime interval=10 secAcceleration  produced=change in velocity/Time=5/10=0.5m/s2Force=mass x acceleration = 2x0.5=1N Id the same force is applied acceleration would remain same.acceleration=change in velocity/timeChange in velocity =acceleration x time=0.5x15=7.5m/sInitial velocity=5m/sFinal velocity = 5+7.5=12.5 m/s  Hope this helps you
```
one year ago
```							M=2kg=t=10secu=5m/secv=10m/secv=u+ata=(v-u)/t=(10-5)/10=1/2m/sec²F=Ma=2×1/2=1kgm/sec²=1Nif this force is applied for 15sec then, velocity isF=mv/tv=(1×15)/2=7.5m/sec.
```
one year ago
```							sorry but the 2nd reply is ½ correct but ½ wrong as u is not =  0 and it is a perception based q ...so u can be 5 or 10 for the 2nd part of q ...giving us ans 12.5m/sst reply> or 22.5 m/s  ...both 12.5 and 22.5 are corrrect
```
one year ago
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