As defined in Chapter 7, the work done by a force on an object is equal to the f

Question

As defined in Chapter 7, the work done by a force on an object is equal to the force times the displacement times the cosine of the angle between the force and displacement vectors (W = F d cos(theta). Suppose you are supporting a 1.77-kg block. What is the gravitational force (magnitude and direction) acting on the block? 17.3 N Incorrect Up Correct: Down If you lower the block a distance of 0.224 m, what is the work done by the gravitational force as you lower the block? 3.89 J What is the value of the angle theta (in degrees - do not enter units)? What is the change in the gravitational potential energy as you lower the block 0.224 m? (If the potential energy increases, the answer should be positive. If the potential energy decreases, the answer should be negative.) -3.89 J Let's switch gears and ask about the electrical force on a charged object. Suppose you have a region where the electric field has a magnitude of 44.2 N/C, the field points straight down, and the field is uniform (i e., the magnitude and direction are the same everywhere in this region). You place an object which has a charge of +0.208 C in this field. What is the magnitude and direction of the electric force acting on the object? 9.19 N Incorrect Up Correct: Down Suppose you lower the object a distance 0.224 m (in the same direction the field is pointing). What is the work done by the electrical force as you lower the block? What is the change in electric potential energy as you lower the object 0.224 m? (If the electric potential energy increases, the answer should be positive. If the electric potential energy decreases, the answer should be negative.)

Explanation / Answer

a)Force on charge =9.19 N which is constant

So Work done by lowering the object a distance 0.224m = 9.19*0.224=2.058 J

b) Change in potential energy=-Work done=-2.058 J

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