Math

Question1. A constant horizontal force on a 200 N is applied to a box in contact with a vertical surface. The coefficient of static friction between the box and the surface is 0.6 , and the coefficient of kinetic friction is 0.4 . Several students are discussing the frictional force on the box 1 second after the force is first applied:
Al : "The frictional force is 60 N since the box will not be moving and the coefficient of static friction is 0.6 ." Brianna: "The frictional force is 100 N upward since the box has a weight of 100 N downward." Carlos: "The frictional force will be 120 N since the box will not be moving and the normal force will be 200 N." David: "The frictional force will be 40 N for the kinetic frictional force and 60 N for the static frictional force. The weight is 100 Nand the coefficient of kinetic friction is 0.4 , giving 40 N for the kinetic friction. Likewise,for the static frictional force it has a coefficient of static friction of 0.6 , giving a static frictional force of 60 N.60 \mathrm{~N} .{ }^{\text {" }}

Studdy Solution
Since the box does not move, the frictional force is equal to the weight of the box, which is 100 N, acting upward to balance the downward gravitational force.
The correct frictional force acting on the box 1 second after the force is applied is:
100N upward \boxed{100 \, \text{N upward}}

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