Question

Two forces oppose one another by acting in opposite directions on a 5 kg object, one being a constant magnitude of 5 N and the other having a variable magnitude of 1/10*^2N. When the object comes to a stop, the second force becomes 5 N as well. How much work is done by the second force to bring the object into static equilibrium?

Answer 1

Answer: The answer is:

0 or none

Step-by-step explanation:

Hope this helps! :)

Answer 2

no work is done by the second force to bring the object into static equilibrium.

Step-by-step explanation:

To find the work done by the second force to bring the object into static equilibrium, we can use the work-energy principle. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.

In this scenario, the object comes to a stop, so its initial kinetic energy (K1) is zero, and its final kinetic energy (K2) is also zero since it's at rest. Therefore, the work done by the second force (W2) is given by:

W2 = K2 - K1 = 0 - 0 = 0

This means that no work is done by the second force to bring the object into static equilibrium. The first force (5 N) alone is responsible for bringing the object to a stop. Once both forces become 5 N and act in opposite directions, their combined effect is to keep the object in static equilibrium, without any net work being done.