Question

a truck starts from the rest and rolles down in a constant acceleration.it travella a distance of 400m in 20s.find the force acting on it if the mass is 7000kg

Answer 1

Given:-

• Initial Velocity of truck = 0m/s

• Distance travelled = 400m

• Time Taken = 20s.

• Mass of truck = 7000kg.

To Find:-

• The force acting on the trunk

Formulae used:-

• S = ut + ½ × a × t²

• F = ma

Where,

S = Distance

u = Initial Velocity

a = Acceleration

t = Time

F = force

m = Mass

Now,

S = ut + ½ × a × t²

400 = 0 × 20 + ½ × a × (20)²

400 = 0 + ½ × a × 400

400 = 0 + 200a

a = 400/200

a = 2m/s²

Hence, The Acceleration is 2m/s²

Therefore,

F = ma

F = 7000 × 2

F = 14000N

Hence, The force acting on the truck is 14000N.

Answer 2

AnswEr :

• Force acting on truck is 14000 N.

GivEn :

• Mass of stone(m) = 7000 kg

• Distance (S) = 400 m

• Time(t) = 20 s

To find :

• Force acting on the truck if the mass is 7000 kg = ?

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How to solve ?

This question can be solved by using the concept of equations of motion and Newton's second law !!

The three equations of motion are,

• v = u + at

• s = ut + (1/2) at²

• v² = u² + 2as

where,

s = displacement

u = initial velocity

v = final velocity

a = acceleration

t = time

Newton's second law is given as,

• F = ma

where,

F = force

m = mass

a = acceleration

Proceeding towards the question..!

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Solution :

First of all we need to find acceleration, as it is unknown.

From the equation of motion,

$\dashrightarrow \sf \: \: \: s = ut + \dfrac{1}{2} a {t}^{2}$

$\dashrightarrow \sf \: \: \: s = 0 \times 20+ \dfrac{1}{2} a \: {20}^{2}$

$\dashrightarrow \sf \: \: \: s = \dfrac{1}{2} \times a \times 20 \times 20$

$\dashrightarrow \sf \: \: \: a = \dfrac{2 \times s}{ {t}^{2} }$

$\dashrightarrow \sf \: \: \: a = \dfrac{2 \times 400}{20 \times 20}$

$\dashrightarrow \sf \: \: \: a = \dfrac{8}{4}$

$\dashrightarrow \sf \: \: \: { \underline{ \boxed{ \bf{ \purple{a = 2 \: m {s}^{ - 2} }}}}} \ \bigstar$

Now,

Force acting on truck ,

$\dashrightarrow \sf \: \: \: F = ma$

$\dashrightarrow \sf \: \: \: F = 7000 \times 2$

$\dashrightarrow \sf \: \: \: { \underline{ \boxed{ \bf{ \orange{F = 14000 \: N}}}}} \ \bigstar$

Therefore, force acting on truck is 14000 N.

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