A constant force acting on a body of mass 5 kg changes its speed from 2 m/s to 8 m/s in 10 s. The magnitude of force is MCQ with Answer and Explanation
A constant force acting on a body of mass 5 kg changes its speed from 2 m/s to 8 m/s in 10 s. The magnitude of force is
A. 5 N
B. 0.5 N
C. 1 N
D. 3 N
Answer: Option D
Solution (By JKExamLibrary)
Acceleration a = (v-u)/t = (8-2)/10 = 0.6 m/s². Force F = m×a = 5 × 0.6 = 3 N. Alternatively, impulse = change in momentum = 5×(8-2) = 30 kg m/s, time = 10 s, so average force = 30/10 = 3 N. Always check direction, but magnitude is 3 N.
Assertion (A): The coefficient of kinetic friction is always strictly less than the coefficient of limiting static friction. Reason (R): Once motion starts, the inertia of rest is broken and interlocking of surface irregularities is less effective.
A.A is true but R is false.
B.Both A and R are true and R is the correct explanation of A.
C.Both A and R are true but R is NOT the correct explanation of A.
Explanation:
Static friction opposes impending motion, reaching a maximum called limiting friction. Once the object moves, it doesn't have enough time for the microscopic irregularities (asperities) of the two surfaces to interlock strongly. Thus, the force required to keep it moving (kinetic friction) is slightly less than the force required to start the motion.
If a stone is dropped from a balloon rising upwards with a velocity of 10 m/s at a height of 75 m, what is the initial velocity of the stone relative to the ground?
Explanation:
Due to the inertia of motion, when an object is dropped from a moving vehicle (or balloon), it inherits the instantaneous velocity of that vehicle. Since the balloon was moving upwards at 10 m/s, the stone's initial velocity relative to the ground is 10 m/s upwards. It will rise slightly before falling.
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