Question

Work done by a body of mass 10kg to lift it through a certain height is 490J . calculate the height

Answer 1

Work done =force*displacement*cos
thetha
As per the information
h=x, force=10*9.8(it is to convert mass into force), work done= 490N, cos thetha=0° which is 1
now putting values in formulae:
490=98*x*1
x=5
The height is 5m.

Answer 2

GIVEN :-

• Mass 10kg is lifted to a certian height
• The work done is 490J

TO FIND :-

• The height to which the body was lifted

SOLUTION :-

We have ,

• m = 10 kg
• W = 490 J
• g = 9.8 m/s²

Work done when mass and height to which the body was lifted is given by ,

$\large {\underline {\boxed {\bigstar {\red{\sf { \: W = mgh}}}}}}$

Where ,

• m is mass of the body
• g is acceleration due to gravity
• h is the height to which the body was lifted

$\implies \sf \: 490 = 10 \times 9.8 \times h \\ \\ \implies \sf \: 490 = 98 \times h \\ \\ \implies \sf \: h = \frac{490}{98} \\ \\ \implies \sf \: h = \frac{ \cancel{490}}{ \cancel{98}} \\ \\ \implies {\underline {\boxed {\blue{ \sf{h = 5 \: m}}}}}$

∴ The height to which the body was lifted is 5 m

ADDITIONAL INFO :-

◉ Work done when the force and displacement are given is given by ,

$\large{\underline{\boxed{\bigstar{\red{\sf{\:W=F.S}}}}}}$

◉ Work done when the power and time are given is given by,

$\large{\underline{\boxed{\bigstar{\red{\sf{\:W=Power\times\:time}}}}}}$

◉ Work done is a scalar quantity. hence it has magnitude but no direction