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
Answers
Given : Initial velocity (u) = 0m/s
Final velocity (v) =
Time = 10 seconds
To Find : Fine Acceleration & Force
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Hence,
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- 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:
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
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\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
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~
Hence,
\therefore\underline{\sf{The~force~acting~on~the~truck~is~\bf{\underline{60000~N}}}}∴
The force acting on the truck is
60000 N
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\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.