Question

A car weighting 500kg working against a resistance of 500 N accelerates from rest to 20m/s in 100 m. What is the work done by the car engine.?

THE CORRECT AND THE DETAILED ANS WILL BE MARKED AS BRAINLIEST...!!!!!​

Answer 1

Given :

▪ Mass of car = 500kg

▪ Frictional force = 500N

(In the opposite direction)

▪ Initial velocity of car = zero

▪ Final velocity of car = 20mps

▪ Distance covered = 100m

To Find :

✒ Work done by the car engine.

Calculation :

⚾ Acceleration of car :

$\implies\bf\:v^2-u^2=2as\\ \\ \implies\sf\:(20)^2-(0)^2=2a(100)\\ \\ \implies\sf\:400=200a \\ \\ \implies\boxed{\bf{\red{a=2\:ms^{-2}}}}$

⚾ Force exerts by engine :

$\mapsto\bf\:F_e=ma\\ \\ \mapsto\sf\:F_e=500\times 2\\ \\ \mapsto\boxed{\bf{\blue{F_e=1000N}}}$

⚾ Net force on the car :

$\implies\bf\:F_{net}=F_e-F_f\\ \\ \implies\sf\:F_{net}=1000-500\\ \\ \implies\boxed{\bf{\green{F_{net}=500N}}}$

⚾ Work done by the car engine :

$\mapsto\bf\:W=F_{net}\:s\\ \\ \mapsto\sf\:W=500\times 100\\ \\ \mapsto\boxed{\bf{\orange{W=50000J=50kJ}}}$

Answer 2

GIVEN :-

Mass of a car, m = 500 kg

Frictional force resisting its motion, f = 500 N

Initial velocity, u = 0 m/s

Final velocity, v = 20 m/s

Distance covered, s = 100 m

TO FIND :-

The work done(W) by the car's engine.

CONSIDER THESE :-

◙ v² - u² = 2as

◙ F = m × a

◙ W = Net force × s

SOLUTION :-

$\bullet$ v = 20 m/s

$\bullet$ u = 0 m/s     (As it starts from rest)

$\bullet$ s = 100 m

v² - u² = 2as

⇒(20)² - (0)² = 2 × a × 100

⇒400 - 0 = 200a

⇒a = 400/200

⇒a = 2 m/s²

$\rule{100}2$

$\bullet$ m = 500 kg

$\bullet$ a = 2 m/s²

F = m × a

⇒F = 500 × 2

⇒F = 1000 N

$\rule{100}2$

$\bullet$ F = 1000 N

$\bullet$ f = 500 N

$\sf{F_{net}=F-f}$

⇒$\sf{F_{net}=1000-500}$

$\underline{\mathbf{\implies F_{net}=500\ N}}$

$\rule{100}2$

$\sf{\bullet\ F_{net}=500\ N}$

$\bullet$ s = 100 m

W = F$\sf{_{net}}$ × s

⇒W = 500 × 100

⇒W = 50000 J (or 50 kJ)             ----→ ANSWER