This question is incomplete, the complete question is;
Flag
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.
Answer:
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
Answer:
787528.7 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).
From the question,
W = Tcos∅(d)............. Equation 1
Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.
But,
d = v(t)..................... equation 2
Where v = velocity, t = time
Substitute equation 2 into equation 1
W = Tcos∅(vt)............. Equation 3
Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s
Substitute into equation 3
W = 210(cos23°)(1.94×2100)
W = 787528.7 J
