Question

a block of mass 8 kg moving with a velocity of 36 km/hr.if its velocity reduced to 18km/hr in 0.5 sec.what is the retarding force on tha block​

Answer 1

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

From the Question,

• Mass of the block,m = 8 Kg

• Inital Velocity,u = 36 Km/h

• Final Velocity,v = 18 Km/h

• Time taken,t = 0.5 s

Since,

Velocity should be given in m/s

Multiplying the magnitude of velocity by 5/18,we get the desired result

Implies,

• u = 10 m/s

• v = 5 m/s

To find

Force acting on the block

From,

$\huge{ \boxed{ \boxed{ \tt{v = u + at}}}}$

Substituting the values,we get :

$\large{ \leadsto \: \tt{5 = 10 + 0.5a}} \\ \\ \large{\leadsto \: \tt{0.5a = - 5}} \\ \\ \large{\leadsto \: \tt{a = - \frac{5}{0.5} }} \\ \\ \huge{ \leadsto \: \underline{ \boxed{ \green{\tt{a = - 10 \: {ms}^{ - 2} }}}}}$

Using the Relation,

$\huge{ \boxed{ \boxed{ \sf{f = ma}}}}$

Substituting the values,we get :

$\large{ \sf{f = ( - 10)(8)}} \\ \\ \huge{ \longrightarrow \: \underline{ \boxed{\sf{f = - 80 N}} }}$

Thus,the force acting on the block is - 80 N

Here,

The block is losing velocity at a rate 10 m/s² due to the action of force.

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

A block of mass 8 kg moving with a velocity of 36 km/hr. If its velocity reduced to 18km/hr in 0.5 sec.What is the retarding force on tha block?

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

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

• Mass of the block (m) = 8 Kg
• Initial Velocity (u) = 36 km/h
• Final Velocity (v) = 18 km/h
• Time Taken (t) = 0.5 Seconds.

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

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Converting the Units,

36 Km/h = ?

$\large{\tt \leadsto 1 \: Km/h = \dfrac{5}{18} \: m/s}$

$\large{\tt \leadsto 36 \: Km/h = 36 \times \dfrac{5}{18} \: m/s}$

$\large{\tt \leadsto 36 \: Km/h = \cancel{36} \times \dfrac{5}{\cancel{18}} \: m/s}$

$\large{\tt \leadsto 36 \: Km/h = 2 \times 5}$

$\large{\leadsto{\underline{\underline{\tt u = 36 \: Km/h = 10 \: m/s}}}}$

18 Km/h = ?

$\large{\tt \leadsto 1 \: Km/h = \dfrac{5}{18} \: m/s}$

$\large{\tt \leadsto 18 \: Km/h = 18 \times \dfrac{5}{18} \: m/s}$

$\large{\tt \leadsto 18 \: Km/h = \cancel{18} \times \dfrac{5}{\cancel{18}} \: m/s}$

$\large{\tt \leadsto 18 \: Km/h = 1 \times 5}$

$\large{\leadsto{\underline{\underline{\tt v = 18 \: Km/h = 5 \: m/s}}}}$

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Applying First kinematics equation,

$\large{\boxed{\tt v = u + at}}$

Substituting the values,

$\large{\tt \leadsto 5 = 10 + a \times 0.5}$

$\large{\tt \leadsto 5 - 10 = a \times 0.5}$

$\large{\tt \leadsto - 5 = a \times 0.5}$

$\large{\tt \leadsto a = \dfrac{- 5}{0.5}}$

$\large{\tt \leadsto a = \dfrac{- 5 \times 10}{5}}$

$\large{\tt \leadsto a = \dfrac{\cancel{- 5} \times 10}{\cancel{5}}}$

$\large{\boxed{\tt a = - 10 \: m/s^2}}$

Therefore, Acceleration of block is - 10 m/s².

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Applying Newton's second law of motion,

$\large{\boxed{\tt F = ma}}$

Substituting the values,

$\large{\tt \leadsto F = 8 \: Kg \times - 10 \: m/s^2}$

$\large{\tt \leadsto F = 8 \times - 10}$

$\huge{\boxed{\boxed{\tt F = - 80 \: N}}}$

So, the Retarding Force on the block is - 80 Newton.

Note:-

• The negative sign on Acceleration & Force indicates that the body is retarding.

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