What force is needed to slide a 250-kg crate across the floor at constant velocity if the coefficient of sliding friction between a crate and a horizontal floor is 0.25?  

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Newton's Second law will help us. It states that the (vector) sum of all forces acting on a body is the same as its mass multiplied by its acceleration, `vecF = m veca.`

There are four forces here: gravitational force `mg` downwards, reaction force `N` upwards, traction force `F_T` horizontally and friction `F_f` also horizontally but in reverse direction.

The acceleration is zero, because velocity is constant. So we obtain

`m vec(g) + vecN + vec(F_T) + vec(F_f) = vec0.`

Vertical forces must balance each other and horizontal, too, so in magnitudes we obtain

`N = mg`  and  `F_T = F_f.`

Also we know that `F_f = mu N,` where `mu` is the coefficient of sliding friction. Hence `F_T = F_f = mu N = mu m g.` This is the final formula, and numerically `F_T approx 0.25*250*9.8 =612.5 (N).`

The answer: the force of about 612.5 N is needed.

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