A constant force vector F = (2,8) moves an object along a straight line from the point (2, 5) to the point (13, 10). Find the work done if the distance is measured in feet and the force is measured in pounds.

Step-by-step explanation:

I will ASSUME that the force vector F = 2i + 8j

which has a magnitude of

F = √(2² + 8²) = √68

which has a direction of

θ = arctan (8/2) = 75.96°

The movement has a distance of

d = √((13 - 2)² + (10 - 5)²) = √146 ft

and a direction of

φ = arctan((10 - 5)/(13 - 2)) = 24.44³

the angle between force and motion is

75.96 - 24.44 = 51.52°

Work is the force in the direction of motion times the distance moved.

W = Fd = (√68)(cos51.52)(√146) = 62 J

or we could just break it into components

x direction work = Fd = 2(13 - 2) = 22 J

y direction work = Fd = 8(10 - 5) = 40 J

total work W = 22 + 40 = 62 J

since the airplane is moving with constant speed then the distance travelled can be calculated by x=u*t

in km/h would be:

x=640*3.5=2.240 km

in m/s would be:

u=640 km/h= 640.000 m/h =640.000/ 3600 m/s =177.8 m/s

t= 3.5 h= 3.5 * 3600 s=12.600 s

then x=u*t= 177.8*12.600= 2.240.280 m

a c

step-by-step explanation: