JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 17)
Consider a force $$\overrightarrow F = - x\widehat i + y\widehat j$$
. The work done
by this force in moving a particle from point
A(1, 0) to B(0, 1) along the line segment is :
(all quantities are in SI units)
_9th_January_Morning_Slot_en_17_1.png)
2
$${1 \over 2}$$
1
$${3 \over 2}$$
Explanation
W = $$\int {\overrightarrow F } .d\overrightarrow r $$
= $$\int {\left( { - x\widehat i + y\widehat j} \right)} .\left( {dx\widehat i + dy\widehat j} \right)$$
= $$ - \int\limits_1^0 {xdx} + \int\limits_0^1 {ydy} $$
= $${1 \over 2} + {1 \over 2}$$ = 1 J
= $$\int {\left( { - x\widehat i + y\widehat j} \right)} .\left( {dx\widehat i + dy\widehat j} \right)$$
= $$ - \int\limits_1^0 {xdx} + \int\limits_0^1 {ydy} $$
= $${1 \over 2} + {1 \over 2}$$ = 1 J
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