A 45.0-kg crate of tools rests on a horizontal floor. You exert a...

Question

A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/s. (c) Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is 1.62 m/s2. (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

Accepted Answer

Hey, everyone in this problem. A 32 kg wooden block is resting on a bench in the lap. A student pulls the block with an increasing horizontal force. The block moves when the pull is greater than 246 newtons. The student notes that the block moves with a constant velocity of 2.1 m per second. When the pull is reduced to 166 newtons, then the block bench system is taken to Mars. OK. So there's a different gravitational acceleration 3.72 m per second squared. And we're asked to determine how much pull will cause the block to move and the acceleration when the pull is 224 noons. OK? And that force accelerates the block at 1.8 m per second squared in the lab. So we're given five answer choices here options a through E each containing a different force and acceleration for this problem. And we're gonna come back to those as we work through this. OK. So let's draw a quick little diagram. We have our block, it is 32 kg and it is resting on a bench in the lab and a student is gonna pull it and we're gonna say that they're gonna pull it towards the right. So if we go ahead and draw a free body diagram of this situation, say that block is resting on a bench, which means that there's gonna be a normal force acting upwards, we have the force of gravity acting downwards. The force of the pull that the student applies is going to be towards the right. Maybe because we've said the block is being pulled to the right. And that means that friction is going to be acting to the left and it's going to oppose the motion. Is there free body diagram? Let's take up into the right to be positive. And what we want to figure out is how much pull will cause the block to move on Mars. So let's start with that question and we are going to use red to indicate no, let's use green to indicate that we're on Mars. OK. So on Mars, if we wanna figure out how much pull is needed, well, what do we have to do in order to get this block moving, we have to overcome static friction. OK. So the minimum force f minimum that we need to pull this block with, in order to get it to move, it's gonna be that maximum static friction force KFS Max, we have to overcome static friction in order for this to move. Now we're called with the static friction force. Fs max, the maximum is gonna be us, the coefficient of static friction multiplied by the normal force on Mars and M. All right. So this coefficient of static friction, we do not know we weren't told in the problem. However, we can use the information about the block in the lab to try to figure that out. OK. So let's call this equation one, so that we can come back to it. We have that our minimum force is going to be equal to that maximum static friction force of us multiplied by N and we're gonna wanna come back to it. OK. Now, the second thing we want to find about Mars is the acceleration when we have a particular force. OK. So let's think about the sum of the forces in the x direction, that direction where that pull force is in the direction where the acceleration will be, that's gonna be equal to the mass multiplied by the acceleration. OK. That's Newton's second law. And again, that acceleration A is what we're looking for. So the acceleration A and then fin from the previous equation, now the sum of the forces in the X direction, let's take a look at our free body diagram. So we have the pull force in the positive direction, the friction force in the negative direction. And at this point in the motion, the block is moving, that means we're gonna be dealing with kinetic friction OK. So we have the pull force FP minus the force of kinetic friction. FK is equal to the mass multiplied by the acceleration. Now, the pull force were given, so we have F pull kinetic friction recall is given by UK multiplied by NF the normal force on Mars and this is equal to the mass multiplied by the acceleration. And we're gonna call this equation no. UK we don't know again, we need to figure out the coefficient of kinetic friction. OK. So we need the coefficient of static friction and the coefficient of kinetic friction from the laboratory experiments because that's gonna be the same on Mars so that we can solve this problem. OK. So let's switch over to blue and blue is going to be on earth. OK. And on earth or in the lab, what do we have? Well, we are to that the block will move if the pull is greater than 246 new. OK. So our maximum static friction force just like a Marx is gonna be equal to mu multiplied by N. But in this case, we have mu S the coefficient of static friction, we have the normal force on earth. All right. So the FS max value we know. OK. That is the force needed to move the block 246 noons. What about this normal force? OK. Well, let's go up and look at our free body diagram and we can calculate this using the forces in the Y direction. OK. So the sum of the forces in the wide direction it's going to be equal to zero. This block is not moving up or down, it's staying at rest on this table. So we have the normal force acting in the positive direction, the force of gravity acting in the negative direction. So N minus FG is equal to zero. So we get that the normal force is going to be equal to the force of gravity, please, which is gonna be equal to the mass multiplied by the acceleration due to gravity. Now, I've written this in blue um but it is going to apply to both Earth and Mars. OK. The difference is gonna be that acceleration due to gravity. So if we go back to our equation or the maximum static friction on earth, OK. We have FS Max is equal to us multiplied by ne well, this is going to be FS Max is equal to us and this normal force is gonna be the mass multiplied by the acceleration due to gravity on earth. All right, substituting in all our values 246 newtons is equal to us multiplied by 32 kg multiplied by 9.8 m per second squared. We're gonna divide both sides by over 32 kg multiplied by 9.8 m per second squared. We get that the coefficient of static virtue in us is gonna be 0.78 44 3878. And we'll leave lots of digits there. So we don't have any round out there. All right. So this is a value that we needed in order to calculate the force that we need to get this block moving on Mars. So now we can go back to equation one and now our minimum force as men, it's going to be equal to this coefficient of static friction. 0.7. Actually, let me leave this as me ask for right now, this coefficient of static friction us multiplied by our normal force. And that's gonna be the same normal force, which is gonna be the mass multiplied by the acceleration due to gravity. But now that G value is gonna be the acceleration due to gravity on Mars G filling this in with our values. Now we have 0.78 443878, multiplied by 32 kg multiplied by this acceleration which we're told is 3.72 m per second squared on Mars working this all out on our calculator. We get that the minimum force needed to get this block moving is gonna be about 93.37 nine dick. No. All right. So we've got the first part of the problem done. Let's take a look at our answer choices. Option A has that this force is 63.1 newtons. Not correct option. B has that. This force is 30.2 newtons. Not correct. So we can eliminate those options. C and D both have 93.4 newtons, which is what we found. So either of those could be correct. Option E is again 63.1 newtons. So that's not correct either. As we've eliminated all of the options except for C and D. Now we need to do part two to figure out which of those is correct. All right. So we got our minimum force. The other thing we were looking for is the acceleration from equation two, we need the coefficient of kinetic friction. So we are gonna go back to the earth just like we did with static friction and figure out the kinetic friction, uh the coefficient of kinetic friction. So let's look at the sum of the forces in the X direction because that's where we have that friction acting. This is gonna be equal to the mass multiplied by the acceleration. Just like in the case on Mars, the sum of the forces in the X direction is gonna be F pulp minus the co uh the kinetic friction force FK. This is equal to the mass multiplied by the acceleration. Now, the situation we were told in the problem, OK is that the block is gonna move with a constant velocity of 2.1 m per second when the pull is reduced to 100 and 66 newtons. Now, if it's moving with a constant velocity, that means the acceleration is zero. So when the force is 100 and 66 newtons, the acceleration is zero and we can substitute that into our equation. OK. So our pull force 166 notes. OK. Minus our coefficient of kinetic friction UK multiplied by the normal force on earth. That's gonna be equal to zero because that acceleration is zero because we have a constant velocity. Now, the normal force on earth, we've already calculated this, this is 32 kg multiplied by 9.8 m per second squared. So we get UK multiplied by 32 kg multiplied by 9.8 m per second squared. It's gonna be equal to 166 newtons. And we're dividing both sides by the 32 multiplied by 9.8. Our units are kilogram meter per second squared, which is equivalent to a newton. So when we divide those out, the units are gonna divide out, we're gonna be left with a unit list value. When we get that in the UK, it's gonna be equal to 0.529 33, 67. All right. So we have our coefficient of kinetic friction and that is going to allow us to go back to equation two and find this acceleration when we have a particular pole. OK. So let's go ahead and substitute our values the pull force. We're interested in is 224 newtons minus this coefficient of friction. We just calculated 0.52933673 multiplied by the normal force. OK. The normal force on Mars just like we did before 32 kg multiplied by the acceleration due to gravity on Mars 3.72 m per second squared. This is gonna equal that mass of 32 kg multiplied by the acceleration that we are looking for. All right, let's simplify as much as we can. So we're gonna divide both sides by 32 kg. Simplify the left hand side. We get that the acceleration is going to be equal to approximately five 0.030867 m per second squared. We had kilogram meters per second squared equivalent to a Newton we divided by kilograms. And so we get those units of acceleration that we want. And that is that acceleration at the particular time when we have a force of 224 noons, going back up to our answer choices. We had narrowed it down to C or D. Option C has that the acceleration is 7 m per second squared. That's not what we found. And so we are left with option D, this has the correct answer for both. We found that the force needed to get the block moving on. Mars is 93.4 newtons and the acceleration when you pull it with a force of 220 new 24 Newtons. Sorry is going to be 5.03 m per second squared. Again, option D thanks everyone for watching. I hope this video helped you in the next one.

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