Question

A shopper pushes her 25 kg grocery cart by a force of 225 N inclined at an angle of 60° with the horizontal through a distance of 7.5 m. Find the total work done by the force applied by the shopper.​

Answer 1

answer is 843.75J

Explanation:

as the force is applied at an angle of 60 ° with the horizontal resolve it into 2 components one x acis fcos60° and other on y axis fsin60°

fsin60° = mg (y axis) as there is no displacement on y axis work done on y axis is zero

on x axis -

work done = fcos60° × displacement

=225*0.5*7.5

=843.75J

Answer 2

■ QuésTion :-

➢ A shopper pushes her 25 kg grocery cart by a force of 225 N inclined at an angle of 60° with the horizontal through a distance of 7.5 m. Find the total work done by the force applied by the shopper.

■ AnsWer :-

➣ 843.75 J

■ Given :-

➤ A shopper pushes her 25 kg grocery cart by a force of 225 N inclined at an angle of 60° with the horizontal through a distance of 7.5 m.

■ SolutiOn :-

A)

$\red{\underline{ \boxed{ \green{\sf \: work = the \: force \: in \: line \: with \: the \: motion \: times \: the \: distance }}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \: Fcos(60)° × 7.5 = \frac{15F}{4} = \color{grey} \boxed{843.75J}$

B) Force of friction :-

$\: \: \: \: \: \: \: \: \: \sf{F_f = (225 × sin(60)° + 9.8 ) × 0.32 }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \longrightarrow\red{\underline{\boxed{\color{yellow}\sf{F_f =140.8} N }}}$

$\sf{F_f}$ is in the opposite direction of motion so the at :

Work of friction = -140.8 × 7.5

= -1056 J

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■ Note :-

➩ Here I give formulae learn that !

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