Question

A truck start ferm rest and rolls down a hill with constant acceleration .it travels a distance of 400 m in 20 s . calculate the acceleration and the force acting on it if its mass is 7 metric tones.​

Answer 1

$\bf{\underline{\underline{Answer:}}}$

∴$\: \bf 14000\: N$

$\bf{\underline {Given:}}$

$⟹Initial\: velocity \:of \:truck\:, u=0$

(since, truck starts from rest)

$⟹ Distance\: travelled, \:s = 400\: m$

$⟹Time, \: t = 20\: s$

$\bf{\underline {To\:Find:}}$

$⟹Acceleration, a=\:?$

$\bf{\underline{\underline{Step\: by\: step \:explanation:}}}$

We know that,

$⟹s=ut+\dfrac{1}{2}at^2$

$⟹400 =0×20+\dfrac{1}{2}×a×{(20)}^{2}$

$⟹400 =0+\dfrac{1}{2}×a×400$

$a=\dfrac{400}{200}=2\:m\:s^{-2}$

Force acting upon truck,

Mass of truck = 7 tonnes = 7 × 1000 kg = 7000 kg

We know that,

Force, $F = m × a$

Therefore, $F=7000×2\:kg\:m\:s^{-2}$

Or$\:\:\:\:\:\:\:\:\:\:F = 14000 \:Newton$

Thus, acceleration of the truck is $\bf 2 \:m\:s^{-2}$ and force acting upon truck in the given condition is 14000 N.

Answer 2

$\huge\underline\mathrm\red{GivEn:-}$

⭐ Initial velocity(u) of truck is 0

⭐ Distance (s) is 400m

⭐ time(t) is 20sec

⭐ mass(m) is 7 metric tonnes

$\huge\underline\mathrm\red{To \: Find:-}$

⭐ Acceleration of truck= ?

⭐ Force acting on it =?

$\huge\underline\mathrm\red{SoLutioN:-}$

To find the acceleration, we use the 2nd equation of motion

$\implies$ S = ut + $\large{\sf {\frac{1}{2}}}$ at²

$\implies$ 400 = 0 x 20 + $\large{\sf {\frac{1}{2}}}$ a(20)²

$\implies$ 400 = $\large{\sf {\frac{1}{2}}}$ a x 400

$\implies$ 400 = a x 200

$\implies$ a = $\large{\sf {\frac{400}{200}}}$

$\implies$ a = 2m/s²

Force acting upon truck,

Mass = 7 tonnes

$\implies$ 7 x 1000 = 7000 kg

Force = mass x acceleration

$\implies$ F = 7000 x 2

$\large{\boxed{\bf{\green{Force = 14000N}}}}$

Thus, the acceleration of truck is 2m/s² and force acting upon truck is 14000 N

$\huge\underline\mathrm\red{ThAnKs...}$