Question: A force of 25N pushes a curling rock from rest 1.0 m across a frictionless surface. Find the FINAL kinetic energy.
attempts:
Well i guess it would be changeinEk
so i tried Wnet=changeinEk
fnet x changeinD = 1/2mvf^2-1/2mvi^2
to fnet x changeinD +1/2mvi^2=1/2mvf^2
Then divided out mass from both sides because it isn't given then cut out vi because it's 0 and isolated vf to get √(2 x fnetxchangeinD=vf
i got vf= 7.07m/s but how am i supposed to find kinetic without mass? No clue about this question, Thanks.
To find the final kinetic energy of the curling rock, you need to know the mass of the rock. However, in this case, the mass is not given directly. But don't worry, there's another way to find the answer.
Since you are given the force (25N) and the displacement (1.0 m), you can use the work-energy principle to solve the problem. The work-energy principle states that the work done on an object is equal to the change in its kinetic energy.
In this case, the force applied is constant, and the displacement is along a straight line. Thus, the work done can be found using the formula: work = force x displacement x cos(theta), where theta is the angle between the force vector and the displacement vector. Since the force and displacement are acting in the same direction (parallel to each other), the angle is 0 degrees, and cos(0) = 1. Therefore, you can simplify the formula to: work = force x displacement.
Given that the force is 25N and the displacement is 1.0m, you can calculate the work done: work = 25N x 1.0m = 25 Joules.
Now, since work is equal to the change in kinetic energy, you can equate it to the difference in kinetic energy between the initial and final states of the curling rock: work = final kinetic energy - initial kinetic energy.
Since the curling rock starts from rest, the initial kinetic energy (KEi) is 0. Therefore, the equation becomes: 25 Joules = final kinetic energy - 0.
Thus, the final kinetic energy (KEf) is 25 Joules.
So, the final kinetic energy of the curling rock is 25 Joules, even though the mass was not provided in the problem.