Question

To keep her dog from running away while she talks to a friend, Susan pulls gently on the dog's leash with a constant force given by fƒ— = (2.3 n )x^ (1.1 n )y^ . How much work does she do on the dog if its displacement is dƒ— = (0.23 m ) y^?

Answer 1

Susan does 0.253 joules of work on her dog by pulling on the leash while the dog's displacement is 0.23 meters in the direction of the force.

Step-by-step explanation:

To calculate the work done on the dog by Susan, we need to use the concept of work in physics. Work is defined as the scalar product (or dot product) of the force vector and the displacement vector. In formula terms, work (W) is F · d, where F is the force applied and d is the displacement.

In Susan's case, the force vector is given as f— = (2.3 N)x^ + (1.1 N)y^ and the displacement vector is d— = (0.23 m)y^. Here, the force has no component in the x-direction concerning the displacement, since the displacement is entirely in the y-direction. Therefore, the work done is simply the product of the magnitude of the force in the y-direction and the magnitude of the displacement.

We calculate the work done as follows:

W = (1.1 N) * (0.23 m) = 0.253 N·m or 0.253 joules, since 1 N·m is equivalent to 1 joule.

Since the force and displacement are in the same direction, the work done is the product of their magnitudes.