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Question

An object of mass 1 kg was moving at a speed of 10 ms1. Constant force acts for 4s on the object and the speed of object was 2 ms1​ in the opposite direction after the applied force was removed. The force acting on the object was


A

The magnitude of force was 3N

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B

The magnitude of force was -3N

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C

The direction of force was same as the direction of initial velocity

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D

The direction of force was opposite to the direction of initial velocity

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Solution

The correct option is D

The direction of force was opposite to the direction of initial velocity


Let initial velocity be u, final velocity be v, and time taken be t.
Initial velocity, u= 10 ms1
Final velocity, v= 2 ms1​ (since this velocity is in opposite direction of the initial velocity which is taken as positive)
Time taken, t= 4s
Let a be the change in acceleration.
a=vuta=2104a=124=3 ms2
From second law of motion, F= 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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