Tactics Box 10.1 Calculating the Work Done by a Constant Force Learning Goal: To
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Question
Tactics Box 10.1 Calculating the Work Done by a Constant Force
Learning Goal:
To practice Tactics Box 10.1 Calculating the work done by a constant force.
To find the work done on an object by a force, begin with a visual overview and follow Tactics Box 10.1.
Recall that the work W done by a constant force F? at an angle ? to the displacement d? is
W=Fdcos?.
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.
Part A
What is the work Ww done on the box by the weight of the box?
Express your answers in joules to two significant figures.
Tactics Box 10.1 Calculating the Work Done by a Constant Force
Learning Goal:
To practice Tactics Box 10.1 Calculating the work done by a constant force.
To find the work done on an object by a force, begin with a visual overview and follow Tactics Box 10.1.
Recall that the work W done by a constant force F? at an angle ? to the displacement d? is
W=Fdcos?.
The vector magnitudes F and d are always positive, so the sign of W is determined entirely by the angle ?between the force and the displacement. TACTICS BOX 10.1 Calculating the work done by a constant force 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 Kdecreases: 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.
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.
Part A
What is the work Ww done on the box by the weight of the box?
Express your answers in joules to two significant figures.
Ww = J Tactics Box 10.1 Calculating the Work Done by a Constant Force Learning Goal: To practice Tactics Box 10.1 Calculating the work done by a constant force. To find the work done on an object by a force, begin with a visual overview and follow Tactics Box 10.1. Recall that the work W done by a constant force F? at an angle ? to the displacement d? is W=Fdcos?. The vector magnitudes F and d are always positive, so the sign of W is determined entirely by the angle ?between the force and the displacement. 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. Part A What is the work Ww done on the box by the weight of the box? Express your answers in joules to two significant figures.Explanation / Answer
Here ,
work done by weight = W*d*cos(90 - 30)
Ww = 2 * 1.8 * cos(60)
Ww = 1.8 J
the work done by weight of the box is 1.8 J
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