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Question

An object of mass 1 kg was moving at a speed of 10ms. A constant force acts for 4s on the object and the speed of object was 2ms in opposite direction after the applied force was removed. The force acting on the object was:


A

3N

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B

-3N

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C

30N

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D

-30N

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Solution

The correct options are
A

3N


D

-30N


Initial velocity u= 10ms
Final velocity v= 2ms (since this velocity is in opposite direction of the initial velocity which is taken as positive)
t = 4s
a=vuta=2104a=124=3ms2F=m×aF=1×(3)=3 Newtons
Here the minus sign indicates that the direction of force was opposite to the direction of initial velocity. Also, the magnitude of force was 3N (magnitude is the absolute value).


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