You are watching: What force do you need to exert to accomplish this?
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This is College Physics Answers with Shaun Dychko. This skier is relocating at constant speed of two meters per second up the slope. They have actually a mass of 75 kilograms and the slope is inclined at three levels. So we"re going to uncover the power output of the skier by dividing the work output separated by the time. The work output will certainly be the force that the skier is, you recognize, exerting on the ground which consequently exerts that force on the skier. So this pressure in other words which I just labeled F multiplied by the distance over which the force is applied, divided by time. Now, we"re not provided distance nor time, but what we are provided is the rate of the skier which is distance over time and so we have the right to replace d over t with v. So, power becomes the pressure that they use multiplied by the rate. So then the trouble becomes how do we discover the force? Well, in the x direction which is along the slope because I"ve tilted the coordinate device so that positive x is up along the slope, that net force in the x direction is the force that the skier applies forward minus the pressure of friction backwards which we"re provided as 25 newlots, and also also minus the component of gravity which is in the negative x direction. That component of gravity F g x is the pressure of gravity multiplied by sine theta bereason in this triangle F g x is the opposite leg of the ideal triangle and this angle up in below is theta. Now, this x component of gravity is zero bereason they"re moving at continuous rate and also so there is no acceleration. So, we have the right to say F equates to force of friction plus mg sine theta where I reput F g through m g and also I moved both of these terms to the best side by adding them to both sides. Then we deserve to relocation every one of this in area of the force F in our power formula. So power output becomes force of friction plus mg sine theta, all times v. So that"s 25 newlots plus 75 kilograms times 9.8 newtons per kilogram time sine of 3, times 2 meters per second which gives 127 watts. Then to find the actual force in component B, well it"s the force of friction plus the component of gravity in the x direction so that"s 25 newtons plus 75 kilograms times 9.8 newloads per kilogram time sine of 3, which provides 63.5 newloads. Then in component C, we"re asked to number out how lengthy it would take to reach a velocity of ten meters per second if they used the exact same force as they were going up the slope on a level surface. So in this case, the net pressure will certainly be the force they used to the ideal, minus the pressure of friction to the left. That"s going to equal mass times acceleration and we"ll divide both sides by m to gain a equals the pressure they applied minus the force of friction separated by their mass. This is the acceleration we can plug into this formula which is that the last speed is the initial speed plus acceleration times time and we"ll solve this for t by subtracting v i from both sides and also then after that dividing both sides by a. We gain time is the last rate minus the initial speed divided by acceleration and also acceleration is this, so we"ll multiply by its reciproca rather of splitting by it. So we"ve v f minus v i times m over F minus F r, so that"s ten meters per second last rate, minus two meters per second initial speed, times 75 kilograms, separated by 63.467 newloads minus 25 newlots, which offers 15.6 secs to reach ten meters per second.
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