Question Friction: A \( 55-\mathrm{kg} \) box rests on a horizontal surface. The coefficient of static friction between the box and the surface is \( 0.30 \), and the coefficient of kinetic friction is \( 0.20 \). A horizontal push force is applied to the box and it starts moving. What must be the magnitude of the push to keep the box moving at a constant speed of \( 0.5 \mathrm{~m} / \mathrm{s} \) ? Express your answer with the appropriate units.

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Transcribed Image Text: Friction: A \( 55-\mathrm{kg} \) box rests on a horizontal surface. The coefficient of static friction between the box and the surface is \( 0.30 \), and the coefficient of kinetic friction is \( 0.20 \). A horizontal push force is applied to the box and it starts moving. What must be the magnitude of the push to keep the box moving at a constant speed of \( 0.5 \mathrm{~m} / \mathrm{s} \) ? Express your answer with the appropriate units.
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Transcribed Image Text: Friction: A \( 55-\mathrm{kg} \) box rests on a horizontal surface. The coefficient of static friction between the box and the surface is \( 0.30 \), and the coefficient of kinetic friction is \( 0.20 \). A horizontal push force is applied to the box and it starts moving. What must be the magnitude of the push to keep the box moving at a constant speed of \( 0.5 \mathrm{~m} / \mathrm{s} \) ? Express your answer with the appropriate units.
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【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2(A) Given mass of box, m=55kg coefficient of stafic frictionmu_(s)=0.3^(0)coefficient of kinesi frictionmu_(k)=0.20As the box is moving.=> kinetic friction aill actThis is the first step of the solution for the given problemExplanation:This is the first step of the solutio ... See the full answer