A 150 g particle at x = 0 is moving at 2.00 m/s in the + x - direction. As it moves, it experiences a force given by Fₓ = (0.250 N) sin (x/2.00 m) . What is the particle's speed when it reaches x = 3.14 m ?

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Hi, everyone. In this practice problem, we're being asked to determine a ball speed at X equals to 5.21 m. If it weighs 200 g, where at X equals zero m, a ball actually moves with a speed of 3.5 m per second along the positive X direction. During the motion, the ball actually encounters an external force following the equation of FX equals 0.75 Newton multiplied by sine of 0.2 X meters to the power of negative one. We're being asked to determine the ball speed at X equals 5. m. And the options given are a 4.4 m per second. B 3.5 m per second, C 5.6 m per second and D 7.8 m per second. So in this problem, we will assume that the given force is the net force acting on the ball. The solution to this problem actually will follow two simple steps. The first one is going to be determining the work done. And the second one, using the work energy principle, we will actually find the ball speed at the given displacement or at X equals to 5.21 m. So first, let's start with uh determining the work done. So the step one is that we will find by integrating the given variable force or integrating the given equation for the external force FX with respect to X to find our work done. So W or work I'm gonna first say finding W so W will equals to in the girl from X knot to X F of FX D X. And substituting our equation for FX given in our problem statement that will then be in the girl from X knot to X F of 0.75 Newton multiplied by sine of 0.2 X to the uh meter to the power of negative one. We want to multiply that by the X as well. So solving for this integral, we will have to actually use the substitution method. And the way for us to do so is to actually by taking Q equals to 0. X meter to the power of negative one, we will actually lu 0.2 and meter to the power of negative one and front so that we will then have U equals to 0.2 m to the power of negative one multiplied that by X. And we will take the derivative of this from the left side and the right side so that we will have D U equals to 0.2 m to the power of negative one multiplied by D X, we will then rearrange this so that we will get one over 0.2 m to the power of negative one multiplied that by D U. And that will actually be equals to D X. And then simplifying this, we will then have five M multiplied by D U or five m D U equals to D X. So the steps in actually solving an integral using D UN D X, especially when we have a sign here actually should follow this following in the integral principles. So you want to recall that the integral of sine A X multiplied by D X will actually equals to the integral of sine of U multiplied by one over a multiplied that by D U, the one over A is just going to be a constant. And according to the principle of integral, we can actually pull that outside of the integral so that this will then be one over a multiplied by the integral of sine of U D U just like. So and the integral of sign U D U is actually just going to be negative co sign of you plus C. The one over A will still remain in front. And in this case, then we will be able to solve integral of sine A X D X. So when integrating with limits, the constant C is going to be dropped here because it cancel out cancels out when substituting the limits. So when we have the integral from X not to X F, we can neglect the C here. So now we can actually substitute uh our in the girl principles that we just noted here and we can rewrite our uh work. So work will then equal to in the girl from X nought to X F of 0.75 Newton multiplied that by sine of 0.2 X meter to the power of negative one D X. And then from here, then uh uh work will then equals to the integral from X knot to X F of open parentheses. 0.75 Newton multiplied by sign of U because we have noted previously that U equals to 0.2 X meter to the power of negative one. Next, we have our D X here that we need to substitute. And we have found that D X actually can be substituted with five MD U. So we want to multiply this by five M multiply that again by D U just like so awesome. So now we have to actually solve this integral by utilizing the integral principles that we just listed here. And that will actually give us open parentheses, 0.75 Newton multiplied by five M. And those two are essentially the constant that we have in this integration. And we want to multiply that by the integral of Sine U which in this case is going to be negative cosine of U. And we want to evaluate all of this from X nought to X F. As I have described previously, the plus C is going to be dropped. In this case, we want to remember that our U is essentially equals to 0.2 X M to the power of negative one. So therefore, in this case, cosine of U can then be uh substituted back with the 0.2 X meter to the power of negative one. So work will then equals two. I'm gonna write down the W 0.75 Newton multiplied by five m multiplied by negative cosine of 0.2 X meter to the power of negative one evaluated from X knot to X F. And in this case, the X knot is given in the problem statement to be zero at X equals zero m and X F is going to be the displacement or the position that we're interested to find at which is 5.21 m just like. So, so we're evaluating from 0 to 5.21 m and calculating this, this will actually get or simplifying this. This will actually turns out to be a negative 3.75 Newton meter multiplied by cosine of 0.2 X meter to the power of negative one still evaluated from zero m to 5.21 m just like. So we want to note that the 0.2 X meter to the power of negative one is dimensionless. So we should use our calculator in order to solve for the cosine here in radiant because radiance is dimensionless. So then calculating or substi or evaluating our co sign, we will then be able to write down our work to be a negative 3.75 Newton meter multiply that by open parenthesis, cosine of 0.2 multiply it by 5.21 m multiplied by a meter to the power of negative one which can be crossed out. We wanna uh minus that with cosine of 0.2 multiplied by zero, which will then just be zero essentially just like that. So we want to substitute our 5.21 and R zero m into our X here. And then that will leave us with the work done to actually be negative 3.75 Newton meter multiply that by 0.50449 minus one, which will come on to a work of 1.858 jules. So that will be the first step of solving this problem which is finding our work done. And then the second step is to employ the work energy principle in order for us to actually find the speed. So from the work energy principle, the total work done or the total work in a system is going to equal to the delta of the kinetic energy or the change in kinetic energy, we know that K or kinetic energy can be calculated by multiplying half of mass multiplied by the velocity squared. So then our work will actually just be the change in kinetic energy or health multiplied by M multiplied by open parenthesis. P F squared minus P zero squared, close parenthesis or essentially the final velocity squared minus the initial velocity squared. What we are interested to look for is the P F and we wanna arrange this equation so that we can find that. So rearranging this, we can get two W divided by M equals to V F squared minus V O squared. Rearranging this further more we will have V F squared to be equals to two W divided by M plus V O squared. So that V F will then just be the square root of two W divided by M plus V O squared just like so awesome. So from the problem statement, we are given that the mass is going to be 200 g or essentially 0.2 kg, we're also given that the V E O is going to be 3.5 m per second. And we have found previously that the work is going to be 1.858 Joles right here, 1.858 Joles. So we have all the necessary information to get V F. So let's substitute all of this information into the expression that we have for V F. So V F is going to be the square root of two W which is then two multiplied by 1.858 chiles divide that by our mass, which is going to be 0.2 kg plus V O squared, which is going to be 3.5 m per second squared. That will give us a V F value of 5.6 m per second, which will actually be the answer to this particular practice problem. So that will actually correspond with option C in our answer choices. So option C with the ball speed at X equals to 5.21 m being 5.6 m per second will actually be the answer to this particular practice problem. So that will be it for this video. If you guys still have any sort of confusion, please make sure to check out our adolescent videos on similar topics and that will be it for this one. Thank you.

A 150 g particle at x = 0 is moving at 2.00 m/s in the + x - direction. As it moves, it experiences a force given by Fₓ = (0.250 N) sin (x/2.00 m) . What is the particle's speed when it reaches x = 3.14 m ?

Verified Solution

Key Concepts

Newton's Second Law of Motion

Work-Energy Principle

Kinetic Energy