Question

A boy pushes a book kept on a table by applying a force of 4.5N Find the work done by the force if the book is displaced through 30 cm along the direction of push.​

Answer 1

AnsWer :-

• Force applied (F) = 4.5 N
• Displacement (s) = 30 cm
• Work done (W) = ?
• cos θ = 0°

The applied force is constant and displacement of the object is also in the same direction along the direction of push, So here we will use the given below formula to calculate the work done :

↠Work Done (W) = Fscos θ

Plug in the known values in above given formula :

↠Work Done (W) = 4.5 N × 30 cm × cos (0°)

↠Work Done (W) = 4.5 N × 30 × 0.01 m × 1

↠Work Done (W) = 4.5 N × 0.3 m × 1

↠Work Done (W) = 4.5 N × 0.3 m × 1

↠Work Done (W) = 1.35 J

Answer 2

$\bf \: Force \: applied \: on \: the \: book \: (F) \: = \: 4.5 \: N$

$\bf \: Displacement (s) = 30 cm$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf= ( \frac{30}{100} ) m = 0.3 m$

$\bf \: Work \: done,$

$\: \: \: \: \: \: \: \: \: \: \: \: \bf \: W = FS$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf= 4.3 \times 0.3$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf= 1.35 J$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \huge \bf \color{teal}(or)$

$\bf \: Given,$

$\bf \: F = 4.5N$

$\bf \: d = 30cm$

$\: \: \: \: \: \bf= 30 \times {10}^{ - 2}$

$\bf \: \: \: \: \: = 0.3m$

$\bf \: So,$

$\bf \: W = F \: . \: d$

$\: \: \: \: \bf \: \: \: \: = 0.3 \times 4.5$

$\: \: \: \: \: \: \: \: \bf = 1.35 \: J$