Question

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

A

The magnitude of force was 3N

B

The magnitude of force was -3N

C

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

D

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

Solution

## The correct options are A The magnitude of force was 3N 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 ms−1​ Final velocity, v= −2 ms−1​ (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=v−ut⇒a=−2−104⇒a=−124=−3 ms−2 From second law of motion, F= m×a⇒F=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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