35 pts please help quick
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.
As we know that spring force is given as
here we know that
F = 4 N
x = 2 cm = 0.02 m
now from the above equation we will have
so the elastic constant of the spring will be 200 N/m
