A force acts on a 8.30 kg mobile object that moves from an initial position of to a final position of in 5.90 s. Find (a) the work done ?

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(a) BYy W = F.∆s
=>W = (4i + 8.5j + 2.7k).(df - di)
=>W = (4i + 8.5j + 2.7k).{(4-3.6)i + (6.7-6.1)j + (5.2-2.9)k}
=>W = (4i + 8.5j + 2.7k).(0.4i + 0.6j + 2.3k)
=>W = 4 x 0.4 + 8.5 x 0.6 + 2.7 x 2.3
=>W = 12.91 J
(b) By P = W/∆t = 12.91/5.9 = 2.19 Watt
(c) By a.b = //a// //b// cosθ 
=>di.df = (3.6i + 6.1j + 2.9k).(4i + 6.7j + 5.2k)
=>di.df = [3.6 x 4 + 6.1 x 6.7 + 2.9 x 5.2) = 70.35
& //di// = √[(3.6)^2 + (6.1)^2 + (2.9)^2] = 7.65
& //df// = √[(4)^2 + (6.7)^2 + (5.2)^2] = 9.38
Thus cosθ = 70.35/[7.65 x 9.38] = 0.98 = cos11.28
=>θ = 11*28'...Show more

a) The displacement is d = df - di.
The work done is then the dot product W= F•d (add component-products)

b) The average power is W/t.

c) Find the dot product P = di•df (add component-products).
Find |di| and |df|.
If angle is θ, then P = |di||df|cosθ
θ = cos⁻¹(P./(|di||df|))

Round all answers to 3 sig. figs and include units....Show more

(a) work = F • s
where • represents the dot product
and F and s are the force and displacement, respectively

s = df - di = [(3.60-4.00) i + (6.10-6.70) j + (2.90-5.20) k] m
s = [-0.40 i - 0.60 j - 2.30 k] m

F•s = (4.00*-0.40 - 8.50*0.60 - 2.70*2.30) J = -12.9 J

(b) avg P = work / time = -12.9J / 5.90s = -2.19 W
but I'm not sure you need the negative sign here. I don't recall seeing "negative power" before.

(c) cosΘ = F•s / |F|*|s|
|F| = √(4.00²+8.50²+2.70²) = 9.77
|s| = √(0.40²+0.60²+2.30²) = 2.41
so
cosΘ = -12.9/(9.77*2.41) = -0.548
Θ = 123º

Hope this helps!...Show more