How Do Kinetic Energy, Work, and Power Interact in Physics Problems?

[nemzy]

I can't seem to get the right answer to these questions

Question:

A 1.8 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x-axis is applied to the block. The force is given by F(x)=(2.4-x^2)i N, where x is in meters and the initial position of the block is x=0

(a) what is the kinetic energy of the block as it basses through x=2.0m?

Answer:

This is how i tried to solve it, since Vi(initial)=0 and Vf(final)= unknown, we can use the formula Vf^2=Vi+2a(X2-X1). But..Force = (2.4-x^2) and at x=2 it is (2.4-4)=-1.6...and f=ma so -1.6/1.8 will equal the acceleration which is -.89...but how the f is the acceleration negative when a horizontal force is applied to the box on a frictionless surface? i don't know where i went wrong with my calculatoins...

Also here is another question that i am having trouble with...

Question:

Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50m/s. At a certain location the conveyor belt moves for 2.0 m u an incline that makes an angle of 12degrees with the horizontal. Assume that the 2.0kg box rides on the belt without slipping

(a) At what rate is the force of the conveyor belt doing work on the box as the box moves up the 12degree incline.

Answer:

this is how i tried to solve it...

since m=2kg, a=0, theta= 12degree, and distance = 2m, and Vc(velocity at constant)=0.50...we have all the constants we need

So first Work=mgdcos(theta) = (2)(9.8)(2)(cos12)=38.343

and X2-X1 (change in distance)=(Vc)(time) sooo...2=38.343t and t = 4 seconds...

and Power = Work/change in time ...sooo 38.343/4 = 9.586 which is not the answer? help what am i doing wrong guys! thanks

 

[deltabourne]

Originally posted by nemzy
I can't seem to get the right answer to these questions

Question:

A 1.8 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an x-axis is applied to the block. The force is given by F(x)=(2.4-x^2)i N, where x is in meters and the initial position of the block is x=0

(a) what is the kinetic energy of the block as it basses through x=2.0m?

Answer:

This is how i tried to solve it, since Vi(initial)=0 and Vf(final)= unknown, we can use the formula Vf^2=Vi+2a(X2-X1). But..Force = (2.4-x^2) and at x=2 it is (2.4-4)=-1.6...and f=ma so -1.6/1.8 will equal the acceleration which is -.89...but how the f is the acceleration negative when a horizontal force is applied to the box on a frictionless surface? i don't know where i went wrong with my calculatoins...

You didn't go wrong anywhere, draw a graph of the force function. It is a negative parabola. Since F=(2.4-x^2), at x= sqrt(2.4) = 1.55m, the force will start pushing back in the negative direction.

Hints:
$$W = \int_{x_i}^{x_f} F(x) dx$$
$$W = \Delta KE$$

 

[M98Ranger]

Hello, it seems like you are on the right track with your calculations. However, there are a few things to keep in mind when solving these types of problems.

For the first question, you correctly identified that the force and acceleration are negative in this case. This is because the force is acting in the opposite direction of the positive x-axis, causing a deceleration of the block. Therefore, the acceleration should also be negative. This doesn't mean that the block is moving backwards, it simply means that its velocity is decreasing.

To solve for the final velocity, you can use the formula Vf^2 = Vi^2 + 2a(x2-x1), where Vi = 0, a = -0.89 m/s^2 (negative because it is in the opposite direction of the force), x1 = 0, and x2 = 2 m. Plugging in these values, you should get Vf = -2.0 m/s. Remember, the negative sign just indicates the direction of the velocity, not the actual value.

For the second question, you correctly used the formula W = mgdcos(theta), but you also need to take into account the work done by the conveyor belt. The belt is doing work on the box by moving it up the incline, so you need to add this work to the work done by gravity. The work done by the belt can be calculated using the formula W = Fd, where F is the force of the belt and d is the distance traveled. Since the belt is moving at a constant speed, the force of the belt is equal to the weight of the box, which is 2 kg x 9.8 m/s^2 = 19.6 N. The distance traveled is given as 2 m, so the work done by the belt is 19.6 N x 2 m = 39.2 J. Adding this to the work done by gravity, you should get a total work of 77.543 J.

To calculate power, you can use the formula P = W/t, where t is the time it takes for the box to travel the distance. You correctly identified that t = 4 s, but you should use the total work of 77.543 J in the formula. This should give you a power of 19.386 W.

I hope this helps clarify the concepts of kinetic energy, work, and power for