Question

. A constant force of 8N acting on an object displaces it through a distance of 3.0 m in the direction of force. Calculate work-done by the force on the object.

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\tt{\therefore{Work\:done\:by\:the\:object=24\:J}}}$

$\sf{Given : } \\ \sf\implies Force \: acting \: on \: object = 8 \: N \\ \\ \sf\implies Displacement \: of \: object = 3 \: m \\ \\ \sf{To \: Find : } \\ \sf \implies Work \: done \: by \: the \: object =?$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\bold{Using \: work \: done \: formula} \\ \sf \implies Work \: done \: by \: the \:object =fs \: cos \: \theta \\ \\ \text{Angle \: between \: force \: and \: displacement} = 0 \degree\\ \\ \sf \implies Work \: done \: by \: the \:object =8 \times 3 \times cos \: 0 \degree \\ \\ \sf \implies Work \: done \: by \: the \:object =24 \times 1 \\ \\ \sf \implies Work \: done \: by \: the \:object =24 \: J$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Work\:done=24\:J}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies Constant \: force(F) = 8 \: N \\ \\ \tt: \implies Displacement(s) = 3 \: m \\ \\ \red{\underline{\bold{To \: Find : }}} \\ \tt: \implies Work \: Done(W.D) = ?$

• According to given question :

$\green{ \star} \tt \: \theta \: = 0 \degree \: \: \: \: (Angle \: between \: force \: and \: displacement) \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Work \: Done = FS \: cos \: \theta \\ \\ \tt: \implies Work \: Done = 8 \times 3 \times cos \:0 \degree \\ \\ \green{ \circ} \tt \: cos \: 0 \degree = 1 \\ \\ \tt: \implies Work \: Done =24 \times 1 \\ \\ \green{\tt: \implies Work \: Done =24 \: J}$