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Grade: 7
        
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

Answers : (3)

Agrata Singh
208 Points
							
Change in velocity=10-5=5m/s
Time interval=10 sec
Acceleration  produced=change in velocity/Time=5/10=0.5m/s2
Force=mass x acceleration = 2x0.5=1N
 
Id the same force is applied acceleration would remain same.
acceleration=change in velocity/time
Change in velocity =acceleration x time=0.5x15=7.5m/s
Initial velocity=5m/s
Final velocity = 5+7.5=12.5 m/s 
 
Hope this helps you 
one year ago
Arun
23047 Points
							
M=2kg=
t=10sec
u=5m/sec
v=10m/sec
v=u+at
a=(v-u)/t=(10-5)/10=1/2m/sec²
F=Ma=2×1/2=1kgm/sec²=1N
if this force is applied for 15sec then, velocity is
F=mv/t
v=(1×15)/2=7.5m/sec.
one year ago
Krutarth
38 Points
							
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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