A block has weight W and mass m, is pulled by tension T at angle θ with the horizon, and experiences a force of friction, F_{k}, opposing its motion across a rough surface. If the box moves a distance d, what is the acceleration of the box?

a) F_{k}d/W

b) (T-W-F_{k})g/W

c) (Tcosθ – F_{k})g/W

d) (Tsinθ – F_{k>})/W

**Explanation**

The acceleration of the box is equal to the sum of forces in the horizontal direction divided by the mass of the box. The sum of horizontal forces is Tcosθ – F_{k}, and the mass in terms of weight is W/g. The distance is unnecessary for the solution.

ΣF_{x}=ma_{x}

Tcosθ – F_{k}=ma_{x}

Tcosθ – F_{k}=W/ga_{x}

a_{x}=(Tcosθ – F_{k})g/W