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Posted by on Monday, November 30, 2009 at 11:11pm.

A box of weight 2N accelerates down a rough plane that is inclined at an angle 30 degrees above the horizontal. The normal force acting on the box has a magnitude 1.7 N, the coefficient of kinetic friction between the box and the plane is O.30, and displacement of the box is 1.8 m down the inclined plane.

Two questions:

A. What is the work done on the box by the normal force?
B. What is the work done on the box by the force of kinetic friction?

Can someone give me the formulas to use?

  • physics - , Monday, November 30, 2009 at 11:13pm

    The work done by normal force is zero, as work is force x distance in the direction of force.

    frictionwork= .30*1.7*1.8 joules

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