Question

The velocity of a body mass 20kg decreases from 20m/s to 5m/sin a distance of 100m .force on the body

Answer 1

$\huge{\underline{\underline{\mathfrak{Answer \colon}}}}$

From the Question,

• Mass,m = 20 Kg

• Initial Velocity,u = 20m/s

• Final Velocity,v = 5m/s

• Distance,s = 100m

To find

Force acting on the body

Using the Relation,

$\huge{ \boxed{ \boxed{ \sf{v {}^{2} = u {}^{2} + 2as }}}}$

Putting the values,we get:

$\sf{5 {}^{2} = 20 {}^{2} + 2(100)a } \\ \\ \implies \: \sf{25 = 400 + 200a} \\ \\ \implies \: \sf{ - 375 = 200a} \\ \\ \implies \: \underline{ \boxed{ \sf{a = - 1.87ms {}^{ - 2} }}}$

We Know that,

F = ma

Putting the values,we get:

$\sf{f = (20)( - 1.87)} \\ \\ \huge{ \sf{f = - 37.4 }}$

Force acting on the body is -37.4 Newtons

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

m = 20 kg.

u = 20 m/s.

v = 5 m/s.

S = 100 m.

$\huge{\bold{\underline{Explanation:-}}}$

Applying third kinematic equation.

$\large{\bold{v^2 - u^2 = 2as}}$

Substituting the values.

$\large{(5)^2 - (20)^2 = 2 \times a \times 100}$

$\large{25 - 400 = 200 \times a}$

$\large{a = \frac{- 375}{200}}$

$\large{\boxed{a = - 1.875 \: m/s^2}}$

Applying Newton's second law.

$\large{\bold{F = ma}}$

Substituting the values.

$\large{F = 20 \times - 1.875}$

$\large{F = - 37.5N}$

Taking only magnitude.

$\huge{\boxed{\boxed{F = 37.5N}}}$

So, the force on body is 37.5N.