A grocery cart with mass of 16 kg is pushed at constant speed ...

Question

A grocery cart with mass of 16 kg is pushed at constant speed up a 12° ramp by a force F_P which acts at an angle of 17° below the horizontal. Find the work done by each of the forces (m $\vec{g}$ , ${\vec{F_{N}}}_{}$ , $\vec{F_{N}}$ ) on the cart if the ramp is 7.5 m long.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A 20 kg crate is being pulled up a 20 degree incline with a constant velocity by a force F at an angle of 15 degrees below the horizontal, determine the work done by the weight mass multiplied by gravity, normal force FN and applied force f on the crate as it travels 10 m along the incline. So that's our end goal is ultimately, we're trying to figure out what the work done by the weight, the normal force and the applied forces on the crate when it travels 10 m along the incline. So that's our final answers that we're trying to solve for. So we're trying to solve for the work for these three separate, OK. These three separate conditions. OK. We're also given some multiple choice answers and the work on the for the grab. So the work so the work done by the weight in this case, in our multiple choice answers, it's denoted as W subscript M and then for the work for the normal force, it's W subscript N and then for the applied force, it's W subscript f respectfully in that order. And all of our answers for all the weight values are in units of jewels. So let's read them off to see what our final answer set might be. A is negative 6.7 multiplied by 10 to the power of 20 and positive 6.7 multiplied by 10 to the power of two B is negative 6.7 multiplied by 10 to the power of 20 and 1.5 multiplied by 10 to the power of two and C is zero and 1.9 multiplied by 10 to the power of two and 1.3 multiplied by 10 to the power of two. And finally, for D it's 1.6 multiplied by 10 to the power of 31.9 multiplied by 10 to the power of two and 1.7 multiplied by 10 to the power of two. OK. So first off, let us create a free body diagram for the crate being pushed up the incline to help us be better visualize this problem to see all the forces that are at work in this particular problem. So I've gone ahead and drawn a diagram to help us visualize all of our forces that are at work in this particular problem. So as we can see, we have our Y and X coordinate. So X is the horizontal and Y is the vertical and we have our right angle right, our triangle here which is a right triangle which denotes our incline. And we have the normal force which is always put, pointing upwards from our crate, which this crate is our little rectangle that's on the incline. And we have our applied force being applied to our crate in this direction, going from the left to the right. So it's going up the incline and from our incline at some angle theta which it says that our crate is being pulled up a 20 degree incline with a constant velocity by a force f at an angle of 15 degrees below the horizontal. So Phi here, which is our Greek symbol here is going to be the angle at which the force is being pushing up at some angle, pushing this crate up the incline. And then from the center of our crate going past down through our incline is we have some angle theta. And then we also have the weight because we have the weight of the box, which is the mass of the box multiplied by the acceleration due to gravity. So you have the mass of the crate, pulling it back down to earth and you also have gravity, pulling it back down to earth. So you have the weight force pulling down on this crate. Ok. So with that all in mind, we now must consider the positive X direction along the incline and the positive Y direction which is perpendicular to the incline. Now, at this point, we need to consider the fact that if the crate is not accelerating, then the net force is zero in all directions. This piece of information can be used to help us find the size of the pushing force. Let us also note that the angles are five equals 15 degrees. And let us also note, we have a little and symbol, let us note that theta equals 20 degrees. Oh Yeah. So let us note that the displacement is in the X direction. And let us also note that the work done by the normal force is zero. Since the normal force is perpendicular to the displacement, the angle between the force of the of gravity and the displacement is, and we can write this as 90 degrees plus fade up equal to 90 degrees plus 20 degrees, which is equal to 110 degrees. And the angle between the normal force and the displacement is 90 degrees. So the angle between the pushing force and the displacement can be written as five plus theta which is equal to 15 degrees plus 20 degrees, which is equal to 35 degrees. So let us note that we can write that the sum of FX is equal to zero. And we can also write that f multiplied by cosine of theta Phi minus M multiplied by G multiplied by sine of theta is equal to zero. So now, at this point, we can rearrange this equation to isolate and solve for F and we can, and when we isolate and sulfur F, we will get that F, the applied force is equal to M multiplied by G multiplied by sine A F all divided by co sign of theta plus P which is equal to. So now we can plug in all of our known variables to solve for the numerical value of F. So when we do that, we know that M the mass is equal to 20 kg multiplied by G which is the acceleration due to gravity, which is 9.8 m per second squared multiplied by sign. So sign of theta which theta is 20 degrees. So it's sign of 20 degrees oh divided by cosine of 35 degrees because we know that theta plus Phi is 35 degrees. So cosine of 35 degrees. So when we take that and we plug it into our calculator, we will find that F is equal to 81.8358 will be round to four decimal places and its units will be Newton's. And let us call this equation I, so we'll just do a little dot dot dot To know that this is to I. So now we must solve for the work done by the weight. So let us recall that the work done by the weight, which we're gonna denote it as W subscript M is equal to M multiplied by G multiplied by D multiplied by cosine of 110 degrees. So now we can plug in all of our known variables to solve for WM, which will give us. So WM. So the work done by the weight is equal to 20 kg multiplied by gravity which is our acceleration due to gravity. So 9.8 m per second squared multiplied for our value of D which D is 10 m in this case is given to us in the prom itself multiplied by cosine of invert the multiplication sign. So cosine of 110 degrees. So our work done our our work due to the weight is equal to negative 670.359 which is approximately equal to negative 6.7 multiplied by 10 to the power of two newtons or sorry jewels, not Newton's jewels. Yes, this is the standard unit of ST unit of measurement for work is in jewels. And this is when we write it in scientific notation, which is important because as we can remember all of our multiple choice answers except for a select few are in scientific notation. So this is our final answer for the work due to the weight. So now we need to solve for the work done by, by the normal force. So now to solve for the work done by the normal force. We need to use the following equation. So WN is equal to FN multiplied by D multiplied by cosine of 90 degrees which is equal to zero joules because as we remember, cosine of 90 degrees is zero. So that's means that our final answer will be equal to zero because any number multiplied by zero is gonna be zero. So that is our final answer for the work due to the normal force. Now, to solve for our final answer, we need to solve for the work done by the applied force F which we can solve for this value by writing the relation, recalling that their relation for the work of the applied force. So the work of the applied force is equal to the applied force multiplied by D where the distance is moved D represents the distance multiplied by cosine of theta plus Phi. So now we could plug in our known variables and solve for WF. So we know that f in this case that we solved together was 81.8. So 85.8358. And don't forget that this unit of measurement is newtons for the force multiplied by 10 m multiplied by. And remember that theta plus Phi was 35 degrees. So multiplied by cosine of 35 degrees. So when we plug that into our calculator, we will find that WF is equal to 670.359 which is approximately equal to 6.7 multiplied by 10 to the power of two jewels. When we, when we convert it to scientific notation, and that is our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter A WM is equal to negative 6.7 multiplied by 10 to the power of two. Joules WN is equal to zero Joules and WF is equal to positive 6.7 multiplied by 10 to the power of two Joules. And that's it we solve for this problem. Thank you so much for watching. Hopefully it helped and I can't wait to see you in the next video. Bye.

A grocery cart with mass of 16 kg is pushed at constant speed up a 12° ramp by a force F_P which acts at an angle of 17° below the horizontal. Find the work done by each of the forces (m $\vec{g}$ , ${\vec{F_{N}}}_{}$ , $\vec{F_{N}}$ ) on the cart if the ramp is 7.5 m long.

Key Concepts

Work Done by a Force

Components of Forces

Gravitational Force and Normal Force

Watch next