Question

a truck of mass 5000kg moving with a speed of 15 m/s comes to rest in 2 seconds what is the force of friction acting on it pls it's urgent

Answer 1

Question:

• A truck of mass 5000kg moving with a speed of 15 m/s comes to rest in 2 seconds what is the force of friction acting on it

Given :

• mass of the truck = 5000 kg
• initial velocity = 15 m /s
• final velocity = 0
• time taken = 2 sec

Solution :

As per the first kinematic equation ,

$\boxed{v = u + at}$

here ,

• v= final velocity
• u = initial velocity
• a = acceleration
• t = time taken

As the truck finally comes to rest v = 0 i.e final velocity = 0

$\implies{u = - at}$

$\implies{a= - \frac{15}{2} }$

$\implies{ a = - 7.5 \frac{m}{ {s}^{2} } }$

Note : here negative sign denotes decceleration

We know that ,

$\implies{F = ma}$

$\implies{F = 5000 \times - 7.5}$

$\boxed{F = - 37500N}$

Answer 2

▪ $\underline\mathrm{Given-}$

• $\mathrm{Mass = 5000Kg}$
• $\mathrm{Initial \: Velocity(u) = 10 m/s}$
• $\mathrm{Final \: Velocity(v) =0m/s}$
• $\mathrm{Time = 2sec}$

▪ $\underline\mathrm{To \: Find-}$

• $\mathrm{Force = ?}$

▪ $\underline\mathrm{Solution-}$

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We Know 1st Equation Of Motion,

★ v = u + at

Placing Values,

$⟶0 = 15 + a \times 2$

$⟶a = \frac{\cancel{-15}}{\cancel{2}}$

$⟶a = </strong><strong>-</strong><strong>7.5 {m</strong><strong>/</strong><strong>s}^{2}$

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We Know Formula For Force,

★ F = ma

Placing Values,

$⟶</strong><strong>F</strong><strong> = ma$

$⟶</strong><strong>F</strong><strong> = 5000 \times - 7.5$

F = -37500 N

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⛬ Force Of Friction is -37500 N