A block of weight w = 15.0N sits on a frictionless inclined plane, which makes a

Question

A block of weight w = 15.0N sits on a frictionless inclined plane, which makes an angle theta = 27.0 with respect to the horizontal, as shown in the figure. (Figure 1) A force of magnitude F = 6.81N , applied parallel to the incline, is just sufficient to pull the block up the plane at constant speed. What is Wg, the work done on the block by the force of gravity w_vec as the block moves a distance L = 5.00m up the incline? What is W_F, the work done on the block by the applied force F_vec as the block moves a distance L = 5.00m up the incline? What is W_N, the work done on the block by the normal force as the block moves a distance L = 5.00m up the inclined plane?

Explanation / Answer

If it moves 4.2 m on incline surface,its height from the ground is H = 4.2 x cos 34* Thus the amount of work done = potential energy = mg H = W x H = work done by applied force ...........the work done by gravity is the component of gravity down the plane x distance, so work by gravity = - m g sin(theta) x distance = - 15N sin34 x 4.2m the minus sign occurs because the force of gravity is in the opposite direction of motion work done by force = 8.39 N x 4.2 m and this will have the same magnitude but opposite direction of the work due to gravity

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