Question

A truck starts from rest and rolls down with a hill with a constant acceleration. it travels a distance of 400m in 20s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint : 1 metric tonne = 1000kg).​

Answer 1

Answer:

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+

2

1

a×t

2

, whereu=0

S=

2

1

a×t

2

a=

t

2

2×S

=

20×20

2×400

=2

sec

2

m

2) Force acting on truck

So,from newton's 2

nd

law of motion

F

=ma=7000×2=14000N=14kN

Explanation:

Hope it helps you

Answer 2

Explanation:

$\frak Given = \begin{cases} \sf{Distance\ travelled,\ s\ =\ 400m.} \\ \sf{Initial\ speed,\ u\ =\ 0\ (it\ starts\ from\ rest).} \\ \sf{Time,\ t\ =\ 20s.} \end{cases}$

To find :- We have to find the acceleration ?

_______________________

$\sf\underline{Distance\ travelled,\ s\ =\ ut\ +\ \dfrac{1}{2}\ at²}\bigstar$

$\frak{\underline{\underline{\dag By\ substituting\ the\ values:-}}}$

$\sf : \implies {400\ =\ 0\ ×\ 20\ +\ \dfrac{1}{2} ×\ a\ ×\ (20)^{2}} \\ \\ \sf : \implies {400\ =\ 200a} \\ \\ \sf : \implies {a\ =\ \dfrac{400}{200}\ =\ 2m/s²} \\ \\ \sf : \implies {\purple{\underline{\fbox{Acceleration,\ a\ =\ 2m/s²}}}}\bigstar$

_______________________

$\frak{\underline{\underline{\dag Now,\ finding\ the\ force\ acting\ on\ it:-}}}$

$\sf : \implies {force,\ F\ =\ m\ \times\ a} \\ \\ \sf : \implies {mass,\ m\ =\ 7\ metric\ tonnes} \\ \\ \sf : \implies {7\ \times\ 1000kg\ (\because 1\ metric\ ton\ =\ 1000kg)} \\ \\ \sf : \implies {=\ 7000kg} \\ \\ \sf : \implies {Force,\ F\ =\ 7000\ \times\ 2} \\ \\ \sf : \implies {\purple{\underline{\fbox{=\ 14000N}}}}\bigstar$

Hence:-

$\sf \therefore {Acceleration\ =\ 2m/s²\ and\ Force\ acting\ on\ it\ =\ 14000N.}$