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All right, hello.
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In this question, we are given information about a block sliding down an inclined plane.
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We have some applied force of 20 newtons pushing the block down, and we have the angle, mass, acceleration, and the initial speed.
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And we're asked for various quantities of work.
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So part a asks, find the work done by the applied force in the first second.
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So we're looking for work by the applied force in the first second.
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Well, we know definitionally work is going to be force dotted with our distance vector, which means we want the parts that are collinear that are together.
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So the force applied times the distance moved times the sine, cosine, cosine of the angle between them.
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And in this case, well, we have our block and it's going to slide just straight down the ramp, which is just in line with the applied force.
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So the cosine of theta, theta is going to be zero, so cosine will just be one in this case.
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So we don't have to worry about that.
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But we do need to find the distance that it travels because we're given that it's going to travel, we want the work done in the first second, but there's no time in this equation.
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So we need to figure out how we're gonna get time.
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Time is gonna translate to distance here.
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And to do that, we're gonna use kinematics.
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Since we know the acceleration, we know the initial speed, we can use the equation x at some t equals our initial position plus v naught times t plus one half at squared.
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In this case, we know the initial speed is zero.
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And we're gonna say that we're just interested in how far it travels.
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So we'll just call our initial position zero.
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And then we know that t is gonna be one second, we know what a is.
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So we're gonna say x after one second is one half times a, which is 10 meters per second squared times one second squared.
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And that will give us five meters.
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So we've traveled five meters in the first second, and now we know our distance.
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So we can go back up here and say work is just our applied force times our distance.
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And that's gonna just equal the 20 newtons times the distance we travel of five meters.
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And that will give us 100 newton meters, also known as 100 joules.
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That'll be our answer for part a.
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Part b asks for the work done by the weight.
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So we want the work done by gravity, essentially.
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So if i go up here and i draw my weight vector of my mass in, i'm gonna have this here.
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And we're looking for the work done by that.
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And we know that work is gonna be force dotted with distance as it was before, which is gonna be fd cosine theta, where theta is the angle between them.
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So we just want the, if i draw my axis like this, positive x this way, we just want the x direction of this weight vector.
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And so to get that, i'm gonna do it in red here.
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I'm just gonna break this vector down into its two components.
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So i have the x component and the y component.
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And in this case, this is gonna be our theta.
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And that's just because of similar triangles, easy just to remember, it's gonna always be on the bottom.
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And so in order to find our, this here, we want force times force, but just the force in the x direction.
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So the force of gravity in the x direction, which is gonna be the force of gravity times the sine of theta times d...