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Question:

How much work does Scott do to push a 76kg sofa 3.3m across the floor at a constant speed?

The coefficient of kinetic friction between the sofa and the floor is 0.25.Wscott= ?explain please!

Answer:

Key phrase: ...at a constant speed... That means no acceleration, so the net force is zero. So Scott's push force is equal in magnitude to the friction force (between the sofa and the floor), but in the opposite direction. Now we need to find the friction force. This is going to involve the relationship between the coefficient of friction (μ), the friction force (Ff) and the normal force (Fn) {The normal force is the force at right angles to the friction surface.} Ff = μ * Fn In this case Fn is just the weight force (mg) of the sofa. Taking (g), the acceleration due to gravity, to be 9.81 m/s? Ff = μ * m * g = (0.25 * 76kg * 9.81 m/s?) = 186.39 N Work done = force * distance Wscott = (186.39 N * 3.3 m) = 615 joules
0 because constant speed makes acceleration 0. F= m X a X displacement

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