At t = 0 a 4.0 kg object is passing the origin, moving right with an initial speed of 1.0 m/s. consider the following two cases: case a: a force f(x) = (1.0n/m) x to the right acts on the object for 4.0 m. case b: a force f(t) = (1.0 n/s) t to the right acts on the object for 4.0 s. what is the change in kinetic energy for the object?

Case A:

Work-energy theorem states that the total work done on an object is equal to the change in the kinetic energy of the object.

Case B:

In this case, we cannot calculate the total work done by , because the force is a function of time, not position.

So, we need to calculate the initial and final kinetic energies.

For the final kinetic energy, we need to find the final velocity. What we have is the force as a function of time. From Newton’s Second Law we can find the acceleration as a function time:

From this function, at t = 4.0 s, v = 2 m/s.

So, the change in the kinetic energy is

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation: