what is the speed of the wire when pulled with 1.3 n ? This is a topic that many people are looking for. thevoltreport.com is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, thevoltreport.com would like to introduce to you A block of mass m = 5.8 kg is pulled up a θ = 25° incline as in Figure P4.30 with a force of magnitu. Following along are instructions in the video below:
Always if you havent read the question yet. Please pause. The video and give give it a look before moving on our first step in this problem as it be with most problems.
Involving forces acting on an object is to draw a freebody diagram of the mass thats situated on the ram since there is no friction in part a of the problem. We do not have a frictional force pointing to the left along this axis here. We simply have the pulling force thats pulling the object up the ramp.
The normal force. Thats acting perpendicular to the surface of the ramp and then the force of gravity. Which of course acts straight down now in most problems in which a box or any other object is on a ramp.
Its going to be helpful to take the fg force and break it into its components. Theres going to be a component along the surface of the ramp and then the component. Thats acting perpendicular to the ramp.
So lets go ahead and do that and when we do that it turns out that the component along the surface of the ramp is f g. Sine. Theta.
The component. Thats perpendicular to the ramp is f g. Cosine.
Theta. And that will always be the case. It might also be helpful to remind ourselves that f g.
Is the same thing as mg. So we can actually replace each fg with mg.
The next step after drawing the free body diagram is to apply newtons second law. Which is the following the sum of the forces is equal to the mass times. The acceleration typically we need to apply newtons second law to both the x and the y direction.
So why dont we try in the x direction. First in fact. Were only going to need to apply this equation in the x direction.
Because in the y direction. It should be pretty clear that the object is not accelerating in other words. The box is not lifting off the surface of the ramp and nor is it digging into the surface of the ramp.
So we do not need to apply newtons second law in the y direction. Because the sum of the forces in that y direction is equal to zero. So we will focus our attention on the x direction and from our free body diagram.
We see that there are two forces acting in the x direction. We have the applied force f. And then we have mg sine theta.
Dont forget that because mg sine theta is pointing leftward that its going to actually be a negative force. So lets fill those two forces into the some of the forces again note. The negative sign and now we can fill in the known values and at this point of course.
We can use our calculators to simplify the left side and then afterwards divide both sides by five point eight. And when doing so the acceleration turns out to be one point three eight meters per second squared. So theres our answer for part a lets turn to part b.
Which is essentially the same as part a except the only difference is that there now is kinetic friction acting between the block and the surface of the incline so that means we just have to add one extra force to our freebody diagram. Because the block is being dragged up the ramp.
The kinetic frictional force will point opposite to that direction down the ramp. So were going to add a force thats acting down the ramp. Weve labeled that force f sub k.
Which simply stands for the kinetic frictional force once again there is no acceleration in the y direction. So we can focus our attention exclusively on the x direction. When we use newtons second law.
We once again have the positively directed applied force its positive because its pointing up the ramp and then these two forces because theyre pointing down the ramp. We can assign a negative sign to them so well go ahead and plug those into the sum of the forces. It turns out that f sub k.
The kinetic frictional force can be rewritten using the following equation the coefficient of kinetic friction multiplied by the normal force we want to note for the normal force that thats actually equal to mg cosine theta. Once again remember there was no acceleration in the y direction. So that means that the normal force is equal in magnitude to mg cosine theta.
They essentially cancel each other out and thats why there is no acceleration in the y direction. So in other words. The normal force can be set equal to mg cosine theta.
So well go ahead and do that we are now ready to substitute in all the known values recall that the coefficient of kinetic. Friction this muse of. K was given to us as 01.
So lets plug in all the known values use your calculator to simplify the left. Side and then divide both sides by. 58 to give the acceleration and you should get a result of 049.
Meters per second squared notice this acceleration is smaller than the one in part a which certainly makes sense because there is friction acting between the block and the ramp and thats kind of slowing down the acceleration of the object. .
Thank you for watching all the articles on the topic A block of mass m = 5.8 kg is pulled up a θ = 25° incline as in Figure P4.30 with a force of magnitu. All shares of thevoltreport.com are very good. We hope you are satisfied with the article. For any questions, please leave a comment below. Hopefully you guys support our website even more.
