Question

A truck starts from rest and rolls downs a hill with constant acceleration it travelles distance of 500 m in 10 sec. find it's acceleration . find the force acting on it if it's mass is 6000 kg pls help me​

Answer 1

Given : Initial velocity (u) = 0m/s

$~~~~~~~~~~~~$Final velocity (v) = ${\sf{\dfrac{500}{10}~=~50m/s}}$

$~~~~~~~~~~~~$Time = 10 seconds

To Find : Fine Acceleration & Force

________________________

$\underline{\frak{As~we~know~that~:}}$

• $\underset{\blue{\sf Acceleration\ Formula}}{\underbrace{\boxed{\frak{\pink{a~=~\dfrac{v~-~u}{t}}}}}}$

$~$

$\qquad\qquad{\sf:\implies{a~=~\dfrac{10~-~0}{10}}}$

$\qquad\qquad{\sf:\implies{a~=~\dfrac{10}{10}}}$

$\qquad\qquad:\implies{\underline{\boxed{\frak{\purple{a~=~10m/s^2}}}}}$★

$~$

$\underline{\frak{As~we~know~that~:}}$

• $\underset{\blue{\sf Force\ Formula}}{\underbrace{\boxed{\frak{\pink{F~=~ma}}}}}$

$~$

$\qquad\qquad{\sf:\implies{F~=~6000~×~10}}$

$\qquad\qquad:\implies{\underline{\boxed{\frak{\pink{F~=~60000}}}}}$★

$~$

Hence,

$\therefore\underline{\sf{The~force~acting~on~the~truck~is~\bf{\underline{60000~N}}}}$

________________________

$\pmb{\frak{More ~Information~:}}$

• Force is defined as the pull or push of an object.

• Acceleration is the change in velocity per unit time.

• Acceleration and force are vector quantity.

Answer 2

Answer:

Given : Initial velocity (u) = 0m/s

~~~~~~~~~~~~ Final velocity (v) = {\sf{\dfrac{500}{10}~=~50m/s}}

10

500

= 50m/s

~~~~~~~~~~~~ Time = 10 seconds

To Find : Fine Acceleration & Force

________________________

\underline{\frak{As~we~know~that~:}}

As we know that :

\underset{\blue{\sf Acceleration\ Formula}}{\underbrace{\boxed{\frak{\pink{a~=~\dfrac{v~-~u}{t}}}}}}

Acceleration Formula

a =

t

v − u

~

\qquad\qquad{\sf:\implies{a~=~\dfrac{10~-~0}{10}}}:⟹a =

10

10 − 0

\qquad\qquad{\sf:\implies{a~=~\dfrac{10}{10}}}:⟹a =

10

10

\qquad\qquad:\implies{\underline{\boxed{\frak{\purple{a~=~10m/s^2}}}}}:⟹

a = 10m/s

2

★

~

\underline{\frak{As~we~know~that~:}}

As we know that :

\underset{\blue{\sf Force\ Formula}}{\underbrace{\boxed{\frak{\pink{F~=~ma}}}}}

Force Formula

F = ma

~

\qquad\qquad{\sf:\implies{F~=~6000~×~10}}:⟹F = 6000 × 10

\qquad\qquad:\implies{\underline{\boxed{\frak{\pink{F~=~60000}}}}}:⟹

F = 60000

★

~

Hence,

\therefore\underline{\sf{The~force~acting~on~the~truck~is~\bf{\underline{60000~N}}}}∴

The force acting on the truck is

60000 N

________________________

\pmb{\frak{More ~Information~:}}

More Information :

More Information :

Force is defined as the pull or push of an object.

Acceleration is the change in velocity per unit time.

Acceleration and force are vector quantity.