An object of mass 1 kg is moving at a speed of 10 m/s. A constant force acts on the object for 4 seconds due to which the speed of object becomes 2 m/s in opposite direction after the force applied is removed. What can be said about the force acting on the object?
An object of mass 1 kg is moving at a speed of 10 m/s. A constant force acts on the object for 4 seconds due to which the speed of object becomes 2 m/s in opposite direction after the force applied is removed. What can be said about the force acting on the object?
The correct options are
A 3N
D -30N
Given, Initial velocity of the object, u= 10 m/s
Final velocity of the object, v= -2 m/s [Negative sign is taken because this velocity is in opposite direction of the initial velocity which is taken as positive]
Time for which force is applied, t= 4 s
Acceleration, a=v−ut [using v=u+at]
⇒a=−2−104
⇒a=−124=−3 m/s2
Now, from Newton's Second Law of Motion, F=m×a
⇒F=1×(−3)=−3 Newton
Here, the minus sign indicates that the direction of force is opposite to the direction of initial velocity and the magnitude of the force is 3 N.
A
3N
D
-30N
Given, Initial velocity of the object, u= 10 m/s
Final velocity of the object, v= -2 m/s [Negative sign is taken because this velocity is in opposite direction of the initial velocity which is taken as positive]
Time for which force is applied, t= 4 s
Acceleration, a=v−ut [using v=u+at]
⇒a=−2−104
⇒a=−124=−3 m/s2
Now, from Newton's Second Law of Motion, F=m×a
⇒F=1×(−3)=−3 Newton
Here, the minus sign indicates that the direction of force is opposite to the direction of initial velocity and the magnitude of the force is 3 N.
