(a) How much work must you do to push a 10 kg block of steel across a steel table at a steady speed of 1.0m/s for 3.0 s? (b) What is your power output while doing so?

Question

Accepted Answer

Hey, everyone. So this problem is dealing with power. Let's see what they're asking us. We are told that food containers are commonly made of aluminum. We have a 20 kg aluminum food crate sliding against an aluminum table that has a kinetic coefficient of friction given by UK equals 0.46. A worker pushes the crate at a uniform speed of 0.80 m/s for five seconds. We are asked to determine the worker's power output. The first thing we can do here is recall, the power is defined by work divided by our change in time. And in turn, work is defined by the force times the displacement or the force multiplied by the displacement. So we can have power equals force multiplied by displacement over delta T. When we draw the free body diagram for the crate, we can determine what forces are acting on it. So we have our pushing force from the worker that's moving that crate in the positive X direction. Our friction force is going to be opposite of the direction of motion. So that'll be the negative X direction. And then we have weight in the negative Y and our normal force in the positive Y. So we were given a delta T from the problem, but we don't yet have our displacement or our force. So let's solve for the force first and then go from there, we will take using Newton's second law, both the X and the Y direction of forces and see if we can solve for this, for this F here. So we have some of the forces in the X direction is equal to mass times acceleration. Again, that's recalling Newton 2nd long for the Y direction, it is going to be the same mass multiplied by the acceleration in the Y direction. Because we are told that the worker pushes the crate at a uniform speed that tells us that acceleration is zero because the derivative of a constant or uniform quantity is zero. So these terms go to zero. And then in the X direction, we have F sub P minus F sub X F equals zero. And we can recall that our friction force is given by UK N. So when we move that friction force to the other side of our equation, we can rewrite that as UK N for our normal force. Now we don't have N but we can go over and look at our Y forces and we will be able to calculate N from there. So we have our normal force acting in the positive direction minus our weight equals zero. And we can recall that weight is given by mass multiplied by gravity. And so when we move weight to that other side of the equation, we can rewrite that as our normal force equals mass multiplied by gravity. And then we can plug that into our pushing force equation. And we are left with UK multiplied by M multiplied by G. From the problem UK was given as 0.46. Our mass was given as 20 kg and gravity is a constant. We can recall 9.8 m/s squared. We plug that into our calculator and we are left with 90.2 newtons. OK. So we found our force and now we need to find our displacement to solve for the power of this worker from our kinematics equations. We can recall that delta X or our displacement is given by V sub I or initial velocity multiplied by time plus 1/2 A for acceleration multiplied by time squared. We've already established that acceleration is zero. So our displacement is going to be equal to our initial velocity, which was given to us in the problem as 0.8m/s multiplied by our time, which was given to us as five seconds. So that equals 4 m. So we now have our force, our displacement and our delta T change in time and we can solve for power. So I'll just rewrite that power equation here that we have determined force multiplied by displacement over delta T. And we will plug that in 90.2 newtons multiplied by 4m divided by five seconds. And we are left with 72.2 watts. So that is the answer for this problem. And that aligns with answer choice. A, that's all we have for this one. We'll see you in the next video.

(a) How much work must you do to push a 10 kg block of steel across a steel table at a steady speed of 1.0m/s for 3.0 s? (b) What is your power output while doing so?

Verified Solution

Key Concepts

Work

Friction

Power