A force F -3 t-5 - N acts on a body whose displacement varies as s -2 t 2-5 - m- Work done by this force in t -0 to 2 sec is inJoule-

Given F = 3t î +5ĵ S = 2t^2 î 5 ĵ By differentiating dSdt=4tî Work done by variable force W = ∫F.dS = ∫_0^2 (3tî + 5ĵ).(4tî) dt W=∫_0^2 12t^2 dt=4(2^3 0^2)=3 ...

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Standard XII

Mathematics

Integration by Substitution

A force F =3 ...

Question

A force $\to F = (3 t^i + 5^j) N$ acts on a body whose displacement varies as $\to s = (2 t^{2}^i - 5^j) m$ . Work done by this force in $t = 0$ to $2 sec$ is (in Joule):

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Solution

Given $\to F = 3 t^i + 5^j$
$\to S = 2 t^{2}^i - 5^j$
By differentiating $\frac{d \to S}{d t} = 4 t^i$

Work done by variable force
$W = \int \to F . d \to S = 2 \int 0 (3 t^i + 5^j) . (4 t^i) d t$
$W = 2 \int 0 12 t^{2} d t = 4 (2^{3} - 0^{2}) = 32 J$

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Q. A force

\to F = (3 t^i + 5^j) N

acts on a body due to which its displacement varies as

\to S = (2 t^{2}^i - 5^j) m .

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Q. A force

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Q. A force

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\to s = (2 t^{2}^i - 5^j)

where

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\to F = (3 t ˆ i + 5 ˆ j)

acts on body due to which its displacement varies as

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A force →F=(3t^i+5^j) N acts on a body whose displacement varies as →s=(2t2^i−5^j) m. Work done by this force in t=0 to 2 sec is (in Joule):

Given →F=3t^i+5^j →S=2t2^i−5^j By differentiating d→Sdt=4t^i Work done by variable force W=∫→F.d→S=2∫0(3t^i+5^j).(4t^i) dt W=2∫012t2 dt=4(23−02)=32 J

A force $\to F = (3 t^i + 5^j) N$ acts on a body whose displacement varies as $\to s = (2 t^{2}^i - 5^j) m$ . Work done by this force in $t = 0$ to $2 sec$ is (in Joule):

Given $\to F = 3 t^i + 5^j$
$\to S = 2 t^{2}^i - 5^j$
By differentiating $\frac{d \to S}{d t} = 4 t^i$

Work done by variable force
$W = \int \to F . d \to S = 2 \int 0 (3 t^i + 5^j) . (4 t^i) d t$
$W = 2 \int 0 12 t^{2} d t = 4 (2^{3} - 0^{2}) = 32 J$