Question

7. A truck starts from rest and rolls down the
hill with Constant acceleration. The truck Covers
400 m in 20 Second .Find its acceleration and
the force acting on it if it's mass is 7 metric
tonnes (1 metric tone=1000kg)​

Answer 1

Given:-

•Initial velocity of the truck(u)=0 (as it starts from rest)

•Distance covered by the truck(s)=400m

•Time taken(t)=20s

•Mass of the truck(m)=7 metric tonnes

=>1 metric tonne=1000kg

=>7 metric tonnes=7×1000

=>7000kg

To find:-

•Acceleration of the truck(a)

•Force acting(F)

Solution:-

By using the 2nd equation of motion,we get:-

=>s=ut+1/2at²

=>400=0×20+1/2×a×(20)²

=>400=200a

=>a=400/200

=>a=2m/s²

Now,from Newton's 2nd law of motion,we get:-

=>F=ma

=>F=7000×2

=>F=14000N

Thus:-

•Acceleration of the truck is 2m/s².

•Force acting on it is 14000N.

Answer 2

• Acceleration = 2 m/s²
• Force = 14 kN

Given

A truck starts from rest and rolls down the  hill with Constant acceleration

The truck Covers  400 m in 20 Second

To Find

Acceleration

Force acting on it if it's mass is 7 metric  tonnes (1 metric tone=1000 kg)

Solution

Initial velocity , u = 0 m/s

[ ∵ start's from rest ]

Distance , s = 400 m

Time , t = 20 s

Apply 2nd equation of motion ,

$\pink{\bigstar}\ \; \bf s=ut+\dfrac{1}{2}at^2\\\\\to \rm 400=(0)(20)+\dfrac{1}{2}a(20)^2\\\\\to \rm 400=0+200a\\\\\to \rm 200a=400\\\\\to \bf a=2\ m/s^2\ \; \pink{\bigstar}$

Now ,

mass , m = 7 metric tonnes

⇒ m = 7 × ( 1000 ) kg

⇒ m = 7000 kg

Acceleration , a = 2 m/s²

Apply Newton Second Law ,

$\pink{\bigstar}\ \; \bf F=ma\\\\\to \rm F=(7000)(2)\\\\\to \rm F=14000\ N\\\\\to \rm F=14 \times 10^3\ N\\\\\to \bf F=14\ kN\ \; \pink{\bigstar}$