A constant force moves an object along the line segment from to . Find the work done if the distance is measured in meters and the force in newtons.

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A constant force F = 6i+8j-6k moves an object along a straight line from point (6, 0, -10) to point (-6, 7, 2).

Find the work done if the distance is measured in meters and the magnitude of the force is measured in newtons.

the work done is -88 J

Explanation:

Given the data in the question;

we know that;

Work done = F × S

where constant force F = ( 6i + 8j - 6k )

S = ( -6i + 7j + 2k ) - ( 6i + 0j - 10k )

S = ( (-6i - 6i) + (7j - 0j) + ( 2k - ( -10k) ) )

S = ( -12I + 7j + 12k )

so

Work force = ( 6i + 8j - 6k ) × ( -12I + 7j + 12k )

Work force = ( 6 × -12 ) + ( 8 × 7 ) + ( -6 × 12 )

Work force = -72 + 56 - 72

Work force = -88 J

Therefore, the work done is -88 J

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

heat the gas and measure the pressure