0.25-kg block oscillates on the end of a spring with a spring constant of 200 N/m. If the oscillations is started by elongating
the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is A :
1 answer:
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Answer:
Frictional force, F = 45.9 N
Explanation:
It is given that,
Weight of the box, W = 150 N
Acceleration,
The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.
It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,
Frictional force is given by :
F = 45.9 N
So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.