TACTICS BOX 10.1 Calculating the work done by a constant force A box of weight w
ID: 1383555 • Letter: T
Question
TACTICS BOX 10.1 Calculating the work done by a constant force
A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ?=30? above the horizontal, as shown (Figure 6) . The normal force acting on the box has a magnitude n=1.7N, the coefficient of kinetic friction between the box and the plane is ?k=0.30, and the displacement d? of the box is 1.8 m down the inclined plane.
What is the work Ww done on the box by the weight of the box?
Direction of force relative to displacement Angles andwork done Sign of W Energy transfer The force is in the direction of motion. (Figure 1) ?cos?W===0?1Fd + The block has its greatest positive acceleration. Kinetic energy K increases the most: Maximum energy is transferred to the system. The component of force parallel to the displacement is less than F. (Figure 2) ?W<=90?Fdcos? + The block has a smaller positive acceleration. K increases less: Moderate energy is transferred to the system. There is no component of force in the direction of motion. (Figure 3) ?cos?W===90?00 0 The block moves at constant speed. There is no change in K: No energy is transferred. The component of force parallel to the displacement is opposite to the motion. (Figure 4) ?W>=90?Fdcos? ? The block slows down, and K decreases: Moderate energy is transferred out of the system. The force is directly opposite to the motion. (Figure 5) ?cos?W===180??1?Fd ? The block has its greatest deceleration. K decreases the most: Maximum energy is transferred out of the system.
Explanation / Answer
The work done by the weight is
Work = (2.0 N) (1.8 m sin 30)
Thus,
Work = 1.8 J [answer]
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