Question

A truck starts from the rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. find its acceleration. Find the force acting on it if its mass is 6 tonnes ( hint: 1 tonne = 1000 kg​

Answer 1

Answer:

Acceleration = 2m/s²

Force acting on the truck = 12kN

Explanation:

u= 0m/s. , s = 400m , t= 20 s

Using 2nd equation of motion

s = ut + 1/2 at²

400 = (0 × 20) + 1/2 × a × 20 × 20

400 = 0 + a/2 × 400

400 = 200a

2 = a

Therefore, acceleration = 2m/s²

According to Newton's 2nd law of motion

Force = mass × acceleration

mass of the truck = 6 tonnes

= (6×1000)= 6000kg

Force acting on the truck = 6000 × 2

= 12000N = 12kN

I hope this will help you :-)

You can verify the answer if you want. :-)

Answer 2

Answer:

Acceleration,a = 2ms^-2

Force,f = 1400N

Explanation:

Given :

Initial velocity of the truck,u => 0 [Since truck starts from rest]

Distance travelled,s => 400m

Time taken,t => 20s

Mass,m => 7 tonnes = (7 × 1000)kg = 7000kg

To Find :

Accelaration,a => ?

Force,f => ?

Solution :

We know that :-

\rm{s=ur+\dfrac{1}{2}at^2

\rightarrow\rm{400m=0\times 20s+\dfrac{1}{2}\times a\times(20)s^2

\rightarrow \rm{400m=0+\dfrac{1}{2}\times a\times400s^{2}

\rightarrow\rm{}400m=\dfrac{1}{1}\times a\times 200s^2→400m=

1

1

×a×200s

2

\rightarrow\rm{400m=a\times200s^2}→400m=a×200s

2

\rightarrow \rm{a=\dfrac{400m}{200s^2}

\rightarrow\rm{a=2ms^{-2}}→a=2ms

−2

Force = m × a

⇒ F = 7000kg × 2ms^-2

⇒ F = 1400N