A constant force F = <2, 8> moves an object along a straight...

A constant force F = <2, 8> moves an object along a straight line from the point (2, 5) to (11, 13). (a) Find the vector x that represents the direction the object is moving. (b) Find F1, the part of F in the direction of x. (c) If the force vector F and displacement vector x are parallel, then the work done by the force F is W and W = |F||x|. How much work is done by our force F in our problem? (Hint: Our force F and displacement vector x are not parallel. What force is being exerted parallel to x?)

Transcript

00:03 In this problem, it is given that a constant force f vector is equal to 2 .8, that is 2 icap plus 8 jcap is acting on an object and the object is displaced from the coordinate 2 .5 to 11 .13 along a straight line.

00:24 So we have to find three parts in this problem and in first part we have to find the displacement vector that is representing the direct.

00:34 Of object is moving.

00:36 So basically here the point a is the start point and b is the end point.

00:47 All right.

00:47 So how do we find the displacement vector? so displacement vector is obtained by subtracting the coordinates, x coordinate and y coordinate and if there is z coordinate then we will also do the same for the z coordinate.

01:01 So we will subtract the starting coordinates from the end coordinates.

01:06 So it will give the displacement vector.

01:09 So x vector will be here the end points are end point in x coordinate is 11 and starting point is 2 therefore it will be 11 minus 2 i cap plus in jcap component the end point is 13 and starting point of the j component is 5 so it is 13 minus 5 j therefore displacement vector that is x vector is 9 i cap plus 8 j cap so this is the vector that is representing the direction of motion of the object so this is the answer for first part now in second part we have to find f1 vector that is the part of the force f vector in the direction of x vector so basically we have to take projection of f vector in the direction of x vector.

02:06 So how do we do that? recall the concept of vector algebra.

02:13 From there, we know that projection of any vector, i'm just first recalling the concept.

02:20 If there are two vectors, a vector and b vector, and we have to find projection of a vector along the b vector.

02:28 So how do we find the projection? so we will write a vector.

02:33 With b vector and this whole thing will multiply the vector along which we have to find the directs along which we have to find the component along b vector so we will write b vector divided by square of modulus of b vector so this will give the projection of a vector along the b vector so we will use this concept here to calculate the f1 vector so how we will write f1 vector f1 vector will be the projection of force vector so f vector along the x direction or the x vector or the displacement vector and displacement vector divided by square of displacement vector now we can put the value of force vectors and displacement vectors here force vector is given as 2 i cap plus 8 j cap we have to take dot product with the x that we have just got in part a, that was 9 icap plus 8 j cap.

03:42 And we have to multiply the entire thing with the x with the x vector, that was 9 icap plus 8 j cap.

03:51 Whole divided by square of the modulus of the x vector...

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