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, Fk, opposing its motion across a rough surface. If the box moves a distance d, what is the acceleration of the box?
a) Fkd/W
b) (T-W-Fk)g/W
c) (Tcosθ – Fk)g/W
d) (Tsinθ – Fk>)/W
Explanation
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 – Fk, and the mass in terms of weight is W/g. The distance is unnecessary for the solution.
ΣFx=max
Tcosθ – Fk=max
Tcosθ – Fk=W/gax
ax=(Tcosθ – Fk)g/W
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