A constant force F= 2i + 3j +4k is applied on a body what will be the work done to move a body 5m in z- direction?

Explanation

Work is equal to the force in the direction of movement (or specified) times the distance moved. Here, the force can be found by using pythagoras to get the magnitude of the force. Using the x(i) and z(k) vectors, the horizontal component can be found, and using the y(j) vector and the horizontal component, the magnitude (hypotenuse can be found),. The angle between the force and the z axis (direction specified) can be found by the taking the inverse sin of the hypotenuse and z(k) vector. Then the magnitude (square root 29) can be multiplied by the sin of the angle ( 47.97) to get 4. This can be multiplied by the distance moved (5m) to get 20J. The work done by a force on an object is given by the dot product of the force and the displacement of the object. In this case, the force is given by F = 2i + 3j + 4k, and the displacement is 5k (in the z-direction). The dot product of these two vectors is given by: F · d = (2i + 3j + 4k) · (0i + 0j + 5k) = 0i + 0j + 20k = 20kTherefore, the work done by the force on the object is 20 times the unit of work (joule) since the force and displacement are in the same direction.