$300$ Joule of work is done in sliding up a $2 \,kg$ block on an inclined plane to a height of $10\, metres$. Taking value of acceleration due to gravity $‘g’ $ to be $10 \,m/s^2$, work done against friction is ........ $J$
$300$ Joule of work is done in sliding up a $2 \,kg$ block on an inclined plane to a height of $10\, metres$. Taking value of acceleration due to gravity $‘g’ $ to be $10 \,m/s^2$, work done against friction is ........ $J$
$100$
$200$
$300$
$0$
$300$ Joule of work is done in sliding up a $2 \,kg$ block on an inclined plane to a height of $10\, metres$. Taking value of acceleration due to gravity $‘g’ $ to be $10 \,m/s^2$, work done against friction is ........ $J$
Work done against gravity$ = mgh$ $ = 2 \times 10 \times 10 = 200\;J$
Work done against friction = (Total work done -work done against gravity)$ = 300 - 200 = 100\,J$
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