Question

Let the vertical axis be the y-axis, with the force of gravity
pointing in the y direction. If a pen with a mass of 200 g is lifted
from the (x; y) point (0.3 m; 0.5 m) to the point (0.75 m; 2 m),
what is the work done by the force of gravity??

Answer 1

Work done by moving from A to B is given by \[ W = \int\limits_{A}^{B} {\bar F} . \delta {\bar l }\]

Where F is the force you're 'pushing' against (a vector) and dl (also a vector) is a small part of the path you're moving along. The integral sign (big S on the left) means that you sum all of these components along the path from A to B. The dot inbetween F and dl means that if the two vectors are at right-angles then that part of the the sum is zero. This means that if the force is along the y-axis (i.e. gravity in this question) displacements in the x-axis don't contribute to the work done.

So, we can completely ignore changes in the x co-ordinate for this problem.

Because F is constant along the path (F=mg, g=9.8N/kg), we also don't actually have to do this sum properly, and can simply write

$$ W = F \delta y = mg \delta y $$

Where dy is the distance moved along the y-axis. So the answer is 0.2kg * 9.81 N/kg * 1.5m = 2.9 Nm = 2.9J

Remember to check your units! If you quote the Earth's gravity in N/kg, you must also convert the mass of the pen to kg, not g.