
Calculate the work done $A$ car of weight $20000\,N$ climbs up a hill at $8\,m/s$ speed.Gaining a height $120\,m$ in $100\,s$.
Answer
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Hint:Work done is the product of force in the direction of motion and the displacement in the direction of motion.Work is done by the force acting on an object which moves some distance.
Complete step by step answer:
Work done can be positive or negative depending on the condition, when work has been done along the displacement then, work done is positive but work has been done against the displacement then, work done is negative. When A car moves upward on the hill at speed $8m/s$ Then gravitational force acts on a car downward. It means that the work has been done by the car against gravity.Then angle between the force and the displacement become ${{180}^{\circ }}$.We have a formula to calculate the work done
$W=F\cdot s$
$\Rightarrow W=Fs\cos \theta $
where, $W=$work done by force, $F=$Force in the direction of motion, $s=$Displacement in the direction of force and $\theta =$Angle between the force and displacement.
S.I unit of work done is Joule ($J$)
Given, $F=20000N$, $s=120m$ , $\theta ={{180}^{\circ }}$
We put these value in the formula of work done
\[W=F\times s\times \cos \theta \]
$W=20000\times 120\times \cos {{180}^{\circ }}$
We put the value of $\cos {{180}^{\circ }}$ in the above equation.
where, $\cos {{180}^{\circ }}=-1$
$W=20000\times 120\times (-1)$
$\Rightarrow W=(12\times 2)\times {{10}^{5}}\,J$
After using the concept of multiplication, we get
$W=24\times {{10}^{5}}J$
Then we divide and multiply by 10
$W=\frac{\text{24}}{\text{10}}\times {{10}^{5}}\times {{10}^{{}}}J$
$\therefore W=2.4\times {{10}^{6}}J$
Hence, the work done by car is $2\times 4\times {{10}^{6}}\,joule$
Note:The negative sign shows the work done by the car against gravity because when the car climbs upward then its weight acts downward and displacement of the car is upward. It means both directions are opposite to each other.
Complete step by step answer:
Work done can be positive or negative depending on the condition, when work has been done along the displacement then, work done is positive but work has been done against the displacement then, work done is negative. When A car moves upward on the hill at speed $8m/s$ Then gravitational force acts on a car downward. It means that the work has been done by the car against gravity.Then angle between the force and the displacement become ${{180}^{\circ }}$.We have a formula to calculate the work done
$W=F\cdot s$
$\Rightarrow W=Fs\cos \theta $
where, $W=$work done by force, $F=$Force in the direction of motion, $s=$Displacement in the direction of force and $\theta =$Angle between the force and displacement.
S.I unit of work done is Joule ($J$)
Given, $F=20000N$, $s=120m$ , $\theta ={{180}^{\circ }}$
We put these value in the formula of work done
\[W=F\times s\times \cos \theta \]
$W=20000\times 120\times \cos {{180}^{\circ }}$
We put the value of $\cos {{180}^{\circ }}$ in the above equation.
where, $\cos {{180}^{\circ }}=-1$
$W=20000\times 120\times (-1)$
$\Rightarrow W=(12\times 2)\times {{10}^{5}}\,J$
After using the concept of multiplication, we get
$W=24\times {{10}^{5}}J$
Then we divide and multiply by 10
$W=\frac{\text{24}}{\text{10}}\times {{10}^{5}}\times {{10}^{{}}}J$
$\therefore W=2.4\times {{10}^{6}}J$
Hence, the work done by car is $2\times 4\times {{10}^{6}}\,joule$
Note:The negative sign shows the work done by the car against gravity because when the car climbs upward then its weight acts downward and displacement of the car is upward. It means both directions are opposite to each other.
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