# SAT Subject Physics Practice Question 27: Answer and Explanation

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**Question: 27**

**4.** An object of mass *m* is traveling at constant speed *v* in a circular path of radius *r*. How much work is done by the centripetal force during one half of a revolution?

A. π*mv*^{2}

B. 2π*mv*^{2}

C. 0

D. π*mv*^{2}*r*

E. 2π*mv*^{2}*r*

**Correct Answer:** C

**Explanation:**

Since the centripetal force always points along a radius toward the center of the circle, and the velocity of the object is always tangent to the circle (and thus perpendicular to the radius), the work done by the centripetal force is zero. Alternatively, since the object’s speed remains constant, the work–energy theorem tells us that no work is being performed.