A 45 g bug is hovering in the air. A gust of wind exerts a force ...

Question

A 45 g bug is hovering in the air. A gust of wind exerts a force F (→ above F) = (4.0î-6.0ĵ) x 10¯² N on the bug. (b) What is the bug's speed at the end of this displacement? Assume that the speed is due entirely to the wind.

Accepted Answer

Hi, everyone in this practice video, we are going to cover this practice problem asking us to determine the magnitude of the ball's horizontal velocity where a 58 g tennis ball is shot outward into the air and there will be a breeze exerting a two D horizontal force with an equation of F equals three times 10 to the power of minus two, I minus five, multiplied by 10 to the power of minus two J Newton on the ball. And the ball actually makes a displacement of 1.5 I minus one J meter. And we're asked to determine the magnitude of the ball's horizontal velocity after the displacement assuming that the vertical velocity has no effect on the horizontal one and the breeze is the only source of the horizontal speed. So the way we wanna tackle for uh this problem is to by first listing all the given information. So first M of the ball is going to equal to 58 g, which will equals to 58 times 10 to the power of minus three kg. And then we can assume that the ball actually starts from rest. So P I will equals to zero m per second. And then the breeze exerts a two D horizontal force with an equation of F equals three times 10 to the power of minus two, I minus five times 10 to the power of minus two J I like to just distribute the times to the power of minus two into each component to make it easier for us to actually see it and just do not forget the to the power of minus two part. And then the displacement I'm gonna call that delta R, the displacement is going to follow 1.5 I minus one J meter just like. So OK. So we're asked to determine the mati of the ball's horizontal velocity after the displacement, we can actually utilize the work energy theorem which is W will equals to the delta or the change in the kinetic energy or delta K. The change in kinetic energy or delta K will equals to half times M V F squared minus half times M V I squared. In this case, because R V I is zero or it starts from rest, we can conclude that the W is essentially just half M V F squared just like. So OK. So next, what we are being asked is the horizontal velocity after this, the displacement or V F. So rearranging this, we can get V F to equal to the square root of two W over M. So we know what the mass of the tennis ball is, but we're not sure what the work is. So we have to solve for the work first before solving four V F. So we recall that the work can be calculated by multiplying the force with the displacement or in this case, because we have uh two D, two D force and two D displacement. We're having, we have to use the dot product. So the work will equals to the dot product of the force and the displacement, the force is three times 10 to the power of minus two in the, in the I component minus five times 10 to the power of minus two in the J component. Newton do with 1.5 in the I component minus one J meter. So we wanna recall uh on how to actually do the dot product, which is essentially um multiplying the I components and um Adding the J components onto that. So this will then be three times 10 to the power of - multiplied by 1.5 in uh which is all of the I components multiplied together. And this will be added with everything in the J component which is minus five times 10 to the power of minus two, Multiplied by -1 just like so and Newton will be multiplied by meter which will equals to Joel. So the we can then be calculated to be 0.45 plus 0.5 G, which will equals to 0.95. So now that we find what the word is, we can actually uh plug that value to find what is being asked in our problem statement, which is the magnitude of the horizontal velocity after the displacement. So the P F is equals to the square root of two W over M and that will be equals to the square root of two multiplied by 0. Divided by the mass which is times 10 to the power of -3 kg. And that will actually equals to 1. m/s. So 1.8 m/s is going to be the magnitude of the ball's horizontal velocity after the displacement which will equals to option B in our answer choices. So option B is going to be the answer to our problem statement. And if you guys still have any sort of confusion on this, please make sure to check out our other lesson videos on similar topics and that'll be all for this one. Thank you.

A 45 g bug is hovering in the air. A gust of wind exerts a force F (→ above F) = (4.0î-6.0ĵ) x 10¯² N on the bug. (b) What is the bug's speed at the end of this displacement? Assume that the speed is due entirely to the wind.

Key Concepts

Newton's Second Law of Motion

Kinematics

Vector Addition

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