MCAT Physics Question — Acceleration
- Sep 15, 2014
- MCAT Question of the Day
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?
c) (Tcosθ – Fk)g/W
d) (Tsinθ – Fk>)/W
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.
Tcosθ – Fk=max
Tcosθ – Fk=W/gax
ax=(Tcosθ – Fk)g/W
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