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 f⃗ at an angle θ to the displacement d⃗ is w=fdcosθ. the vector magnitudes f and d are always positive, so the sign of w is determined entirely by the angle θ between the force and the displacement.

Asmall, positively charged ball is moved close to a large, positively charged ball. which describes how the small ball likely responds when it is released? it will move toward the large ball because like charges repel. it will move toward the large ball because like charges attract. it will move away from the large ball because like charges repel. it will move away from the large ball because like charges attract.

Next a skier is pulled by a tow rope up a frictionless ski slope that makes an angle of 15 with the horizontal. the rope moves parallel to the slope with a constant speed of 0.69 m/s. the force of the rope does 800 3 of work on the skier as the skier moves a distance of 8.4 m up the incline. (a) if the rope moved with a constant speed of 2.2 m/s how much work would the force of the rope do on the skier as the skler moved a distance of 8.4 m up the incline? at what rate is the force of the rope doing work on the skier when the rope moves with a speed of (b) 0.69 m/s and (c) 2.2 m/s?

The mass of the sun is 1.99×1030kg and its distance to the earth is 1.50×1011m. what is the gravitational force of the sun on the earth?