Minimum Force for Overcoming Friction on a Wooden Box

Thread starter JoeDoe
Start date Mar 28, 2011
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Force

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To determine the minimum force required to move a 20 kg wooden box on a horizontal stone floor, the coefficient of kinetic friction is essential. The force of friction can be calculated using the formula F_friction = μ * N, where μ is the coefficient of kinetic friction (0.40) and N is the normal force (equal to the weight of the box). The weight of the box is 20 kg multiplied by the acceleration due to gravity (approximately 9.81 m/s²), resulting in a normal force of 196.2 N. Therefore, the minimum force needed to overcome friction is 0.40 * 196.2 N, which equals 78.48 N. This calculated force is the minimum required to initiate movement of the box on the stone floor.

Mar 28, 2011

JoeDoe

Homework Statement

A 20 kg wooden box is pushed a distance of 15 meters on a horizontal stone floor, by a force just sufficient to overcome the friction between the box and floor. The coefficient of kinetic energy is 0.40. What is the minimun force required?

Thanks.

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Mar 28, 2011

SammyS

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