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Work Done When Force Is Varying

A force f bar...

Question

A force f(bar)=5i(cap)–3j(cap)+2k(cap) moves a particle from r1(bar)=2i+7j+4k to r2=5i+2j+8k. Calculate the work done

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Solution

Work done=f.(r2-r1)
r2-r1=(5i+2j+8k) -(2i+7j+4k)
=3i-5j+4k
F.(r2-r1)=(5i-3j+2k).(3i-5j+4k)
(53)+(-3-5)+(2*4)
=15+15+8
=38 J

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Similar questions

F = 5 i - 3 j + 2^k

r_{1} = 2 i + 7 j + 4^k

r_{2} = 5^i + 2^j + 8^k \cdot

{\to r}_{1} = 3^i + 2^j - 6^k

{\to r}_{2} = 14^i + 13^j - 9^k

(4^i +^j + 3^k)

N

A particle moves from position vector r1=3i^+2j^-6k^ to position vector r2=14i^+13j^+9k^ under the action of force 4i^+j^ +3k^N.Find the work done.

(4 i + j + 3 k) N

r_{1} = 3 i + 2 j - 6 k

r_{2} = 14 i + 13 j + 9 k,

P = 2 i - 5 j + 6 k

Q = - i + 2 j - k

4 i - 3 j - 2 k

6 i + j - 3 k .

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