CALC A 2.6 kg block is attached to a horizontal rope that exerts a variable force Fx=(20−5x) N, where x is in m. The coefficient of kinetic friction between the block and the floor is 0.25. Initially the block is at rest at x=0 m . What is the block's speed when it has been pulled to x=4.0 m?

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Hi everyone in this practice problem. We are being asked to determine the speed of a box when it is pushed to a distance of five m. If the box is initially at rest, we will have a laborer pushing a 3.2 kg box on level ground with a force given by FX equals to 18 minus three X Newton where X is going to be the distance or the position in meter. We wanna consider that the UK between the box and the ground is 0.23 we are being asked to determine the speed of the box when it is pushed to a distance of five m, which is initially going to be at rest. The options given are a 3.2 m per second. B 10.2 m per second, C 5.73 m per second and D 3.61 m per second. So we will consider the box to be a particle and I am going to start us off with giving you guys a pictorial representation or diagram or a free body diagram for our system. So we will have the ground here and we'll have a box initially at rest, the box itself is going to be experiencing the force from the labor, pushing the box to the right. And it will actually be also experiencing in red, a friction force counter to that. So the friction force is going to be F K with a UK of 0.23 given and then the force pushing force is going to just be F and that will be the two forces acting on the horizontal axis. However, on the vertical axis, it will experience first the weight of the actual uh box itself, which is just going to be M multiplied by G. And at the same time, the normal force from the box and the ground pointing upwards in the vertical direction and both the normal and the weight are going to be the two forces acting on the vertical direction. Ok. So initially the box will be positioned as X knot or X zero equals to zero m and it will have the speed of the starting speed which is zero m per second because it will be starting from rest. However, after the laborer pushed the box, it will then B position at X one equals five m and V one is actually going to be the speed of the box that we are going to be finding out, which is the one that we are gonna be interested to look at the forces exerted onto or by the box is going to still be the same. So essentially on the horizontal axis, there will be the force of the labor pushing and uh uh friction force countering that and then going there is also going to be the normal force and the weight on the vertical direction as well. Awesome. So now we wanna actually first apply Newton's second law in the right direction in order for us to be able to know what the normal force is, because we will need that normal force to then for us to be able to calculate the frictional force between the book and the ground. So Newton's second law in the right direction. According to Newton's second law, the Sigma of all the forces acting on the Y direction or Sigma F Y will be equals to M multiplied by a if the box is moving in the vertical direction. However, because the box is staying still in the vertical direction, then Sigma F Y is then going to be equals to zero Newton or zero. The two forces acting on the Y direction is going to be N minus that by the weight equals to zero. So the normal force will then equals to the weight which is going to equals to M multiplied by G just like so, so then our friction force will then equals to F K equals to mu K multiplied by the normal force N which will be mu K multiplied by M multiplied by G Awesome. So we want to hold on to this for a while. And what we want to do in the end to find the speed of the box is to basically use the energy principle which is going to be the network done. So according or calculating that work done W net or the network done will then be equals to the change in kinetic energy or delta K plus the change in thermal energy or delta E T H just like. So, and what we wanna be able to do with this formula is to then actually um open up the delta K to then get the final speed of the box. So before we can do that, we want to calculate the total work done on the box and also the change in the thermal energy which we can find by using integration. So first uh W net will equals to the integral of F net multiplied by D X looking at the horizontal position which is where the change is happening. We will integrate um F net starting from position zero which is X NOTT zero m to position X one which is five m. So we were integrating F net D X from X nought to X one which is from 0 to 5 m. And the F net is actually given in our problem statement which is going to be FX equals to A P minus three X Newton. So we wanna substitute that for our F net, which in this case, then our equation will then be the integral from 0 to m of A P minus three X multiplied by D X. And we want to recall that the integral of a constant D X will then be C X and the integral of X multiplied by ad X is going to be half X squared. So you want to substitute that or use these principles to find our in the girl which will give us um the in the girl from so open parenthesis of A P D X which is going to be X minus three X which is going to be 3/2 X squared. And we want to evaluate that integral from 0 to and substituting everything, we will get the W net to be 52.5 Joles just like. So, so we get our W net. What we wanna be calculating next is the delta E P H or the change in internal energy. So the delta E P H will be calculated from the frictional force. So delta E T H equals to the integral from X knot to X one of F K multiplied by the X. And that will be the same, which is the integral from 0 to 5 m of F K which we have calculated earlier or found a formula earlier which is M K multiplied by M multiplied by G. So we want to substitute that here, which is mu K multiplied by M multiplied by G multiplied by D X just like. So, so Mu K multiplied by M multiplied by G is essentially just a constant. So we will follow the integral of CD X equals to C X for this particular integral right here. So this equation will then be open parenthesis of mu K multiplied by M multiplied by G X evaluated from 0 to 5. And that will essentially be delta E P H equals to open parenthesis UK is going to be 0. which is given in the problem statement. Um M is going to be 3. kg which is also given in the problem statement and R G is 9. m per second squared X is still gonna be X which is going to be evaluated from 0 to 5. Substituting everything and calculating delta E P H will then equals to 36.1 jules. So now that we have calculated our delta E P H and also our W net, we can actually uh substitute them back into our network. Done equation right here in order for us to be able to finally calculate uh the speed of our box in the final condition. So then substituting them into this equation rearranging it as well. At the same time, delta K will then equals to double un net minus delta E T H and that will essentially give us 52.5 Joles minus 36. Joles, which will then equals to 16.4 Joles. So delta K will then equals to 16.4 Joles. Awesome. So we want to then um open up our delta K in order for us to find the speed of our box. So delta K equals to 16. jules and delta K is equals to half M V F squared minus half M V I squared. We know that V I is zero because the box initially started at rest. So delta K will then just equals to half M V F squared. So then 16.4 Jule will equals two half multiplied by M which is 3.2 kg, which is the mass of our box and V F squared, which is the uh final speed that we are interested at looking. So calculating this, we can actually get V F to then be equals to 3.20 m per seconds. So through rearranging and taking the square root or V F will then come out to be 3.2 m per second, which will actually correspond to option A in our answer choices. So with the speed of the box, when it is pushed to a distance of five m of 3. m per second, option A is going to be the answer to this particular practice problem. So if you guys still have any sort of confusion on this particular video. Please make sure to check out our other lesson videos on similar topics and that'll be all for this one. Thank you.

CALC A 2.6 kg block is attached to a horizontal rope that exerts a variable force Fx=(20−5x) N, where x is in m. The coefficient of kinetic friction between the block and the floor is 0.25. Initially the block is at rest at x=0 m . What is the block's speed when it has been pulled to x=4.0 m?

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Key Concepts

Newton's Second Law of Motion

Work-Energy Principle

Frictional Force