A block of mass 10 kg is accelerating at 2 m/s^2 on a horizontal surface under an applied force of 50 N. What is the magnitude of the frictional force opposing its motion? MCQ with Answer and Explanation
A block of mass 10 kg is accelerating at 2 m/s^2 on a horizontal surface under an applied force of 50 N. What is the magnitude of the frictional force opposing its motion?
A. 20 N
B. 50 N
C. 70 N
D. 30 N
Answer: Option D
Solution (By JKExamLibrary)
According to Newton's Second Law, the net force on the block is F_net = mass * acceleration = 10 kg * 2 m/s^2 = 20 N. The net force is the difference between the applied forward force and the opposing frictional force (F_net = F_applied - F_friction). So, 20 N = 50 N - F_friction. Solving for friction gives F_friction = 50 - 20 = 30 N.
Raindrops fall to the ground with a uniform, constant velocity. This constant velocity is achieved when the gravitational pull is perfectly balanced by:
Explanation:
As a raindrop falls, it accelerates due to gravity. The upward viscous drag force of the air increases with the drop's velocity (Stokes' Law). Eventually, the downward weight of the drop is exactly balanced by the upward buoyant force of the air plus the upward viscous drag force. The net force becomes zero, and it falls with a constant 'terminal velocity'.
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