Question

Physics Question:- A truckstarts from rest and rolls down a hill with constant acceleration . it travels a distance of 400m in 2s . find its acceleration . Find the force acting on it if its mass is 7 metric tones

Answer 1

$\huge\underline{\overline{\mid{\bold{\red{ANSWER-}}\mid}}}$

Initial velocity, u = 0 (since the truck is initially at rest)

Distance travelled, s = 400 m

Time taken, t = 20 s

According to the second equation of motion:

s=ut+1/2 at'2

Where,

Acceleration = a

400=[0]20+1/2[a] [20] [20]

400=0+1/2[400] [a]

400=[200] [a]

400/200=a

2=a

1 metric tonne = 1000 kg (Given)

∴ 7 metric tonnes = 7000 kg

Mass of truck, m = 7000 kg

From Newton’s second law of motion:

Force, F = Mass × Acceleration

F = ma = 7000 × 2 = 14000 N

Hence, the acceleration of the truck is 2 m/s2 and the force acting on the truck is 14000 N.

Answer 2

GivEn:-

• The truck starts from rest, so initial Velocity = 0 m/s

• Distance travelled by truck = 400m

• Time taken by truck to travel the distance = 20 sec

• Mass of truck = 7 tonne = 7000 kg

To find:-

• Force acting on truck

SoluTion:-

✩ As we know the equation of motion,

★ ${\underline{\boxed{\bf{\purple{S = ut + \dfrac{1}{2}at^2}}}}}$

Here,

• u denotes initial Velocity

• S denotes Distance travelled by truck

• t denotes time taken by truck

• a denotes acceleration

• m denotes mass of truck

• F denotes force

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✮ Now put the givEn values,

$:\implies\sf 400 = 0 \times 20 + \dfrac{1}{ \cancel{2}} \times a \times 20 \times \cancel{20}$

$:\implies\sf 400 = 0 + a \times 20 \times 10$

$:\implies\sf 400 = 200a$

$:\implies\sf a = \cancel{ \dfrac{400}{200}}$

$:\implies{\underline{\boxed{\bf{\pink{a = 2\;ms^{-2}}}}}}$

✮ Now as we know that,

★ ${\underline{\boxed{\bf{\purple{F = ma}}}}}$

Now put the value of m and a in above formula,

$:\implies\sf F = 7000 \times 2$

$:\implies{\underline{\boxed{\bf{\pink{14000\;N}}}}}$

$\therefore\sf\; \underline{Hence,\; Acceleration\;is\;2\;ms^{-2}\;and\;force\;is\;14000\;N.}$

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Additional Information:-

★ There are three equations of motion:-

⠀⠀⠀✩ $\sf v = u + atv=u+at$

⠀⠀⠀✩ $\sf s = ut + \dfrac{1}{2} at^2$

⠀⠀⠀✩ $\sf v^2 - u^2 = 2as$

⠀⠀ ━━━━━━━━━━━━━━━━━━━━━

$\;\;\star\;\sf Acceleration (a) = \dfrac{Final\; velocity (v) - Initial\; velocity (u)}{Time (t)}$