Question

A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20. Find its acceleration. Find the force acting on it if its mass is 7 tonnes.

Answer 1

Given data:

Mass of stone(m)=7000 kg

Distance(S)=400 meter

Time(t)=20 second

1) We have to find the acceleration of the truck

So from the equation of motion S= ut+1/2      at 2 , where u=0

S=1/2aX t2        

 a=    2×S/t2   =   2×400/20×20    =2m/ sec 2

    2) Force acting on truck

So, from newton's 2 nd  law of motion

F  =ma=7000×2=14000N=14kN

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Answer 2

$\huge\star{\underline{\underline{\sf\purple{ Solution:-}}}}$

GivEn:

• Distance covered by the truck (s) = 400m
• Time taken (t) = 20 seconds
• Initial velocity (u) = 0

To Find:

• Acceleration (a) = ?
• Force (f) = ?

$\\ \star{\underline{\sf\purple{ Formula:-}}}$

• $\bullet{\underline{\boxed{\sf{s = ut + \dfrac{1}{2} at^2}}}}$

• $\bullet{\underline{\boxed{\sf{Force = Mass \times Acceleration}}}}$

First Case:

$\bullet{\underline{\boxed{\sf\pink{s = ut + \dfrac{1}{2} at^2}}}}$

After putting Values;

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

$\implies {\sf{ 2 \:ms^{-2} }}$

Acceleration (a) of the truck is 400 m/s².

Second Case:

$\bullet{\underbrace{\underline{\boxed{\sf\pink{Force = Mass \times Acceleration}}}}}$

• Mass of Truck = 7 tonnes
• 7 tonnes = 7000 kg

$\implies {\sf{ F = 7000 \times 2 }}$

$\implies {\sf{ F = 14000 \:ms^{-2} }}$

$\implies {\sf{ F = 14000 \:kg\:ms^{-2} }}$

$\implies {\sf{ 14000 \:N }}$

Hence,

The Force is 14000 N is acting on truck.