A block of weight w = 30.0N sits on a frictionless inclined plane, which makes a

Question

A block of weight w = 30.0N sits on a frictionless inclined plane, which makes an angle theta = 23.0^circ with respect to the horizontal, as shown in the figure. (Figure 1) A force of magnitude F = 11.7N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. What is W_g, the work done on the block by the force of gravity w_vec as the block moves a distance L = 4.30m up the incline? Express your answer numerically in joules. What is W_F, the work done on the block by the applied force F_vec as the block moves a distance L = 4.30m up the incline? Express your answer numerically in joules. What is W_N, the work done on the block by the normal force as the block moves a distance L = 4.30m up the inclined plane? Express your answer numerically in joules.

Explanation / Answer

Wg = m g sin d = 30 * 0.39073 * 4.3 = -50.4 J

WF = F d = 11.7 * 4.3 = 50.3 J

W_N = N d cos90 = 0 J

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