Question

1.A shopper pushes her 25 kg grocery cart by a force of 225 N inclined at an angle of 60° with the horizontal through a distance of 7.5 m. Find the work done by the force applied by the shopper.

2.Horizontal force F applied to an object varies with its displacement. Calculate the work done by the force when the object moves from x=0 to x=10.0 m

Answer 1

1. The horizontal projection of the force is:

$$F_x = 225N\cdot \cos60\degree = 112.5N$$

By definition, the work is:

$$W = F_xs$$

where $s = 7.5m$ is the horizontal displacement. Thus, obtain:

$$W = 112.5N\cdot 7.5m \approx 843.8J$$

2. By the general definition, the work is given as follows:

$$W = \int_{x_0}^xF_xdx$$

where $x_0 = 0$, $x = 10m$ are start and end points, and $F_x$ is the projection of the force on x-axis.

Answer. 1) 843.8 J, 2) $W = \int_{0}^{10}F_xdx$.