Question

a truck starts from rest and roll down a hill it travels a distance of 400m in 20s find its acceleration​. find thr force acting on it if it's mass is 7 tonnes.​

Answer 1

Answer:

★ Acceleration = 2 m/s²

★ Force = 14000 N

Step-by-step explanation:

Given:

• Truck is starting from rest.

So, initial velocity (u) = 0

• Distance (s) = 400 m
• Time (t) = 20 seconds
• Mass = 7 tonnes = 7000 kg

To find:

⇢ Acceleration of the truck = ?

⇢ Force acting on it = ?

Solution:

From Second equation of motion,

$\boldsymbol{S = ut + \dfrac{1}{2}a {t}^{2} }$

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

$\implies \sf{400 = 200a}$

$\implies \sf{a = \dfrac{400}{200} }$

$\therefore \: \boxed{ \bf{Acceleration = 2 \: m /{s}^{2} }}$

Now, for finding force, we know that:

$\boldsymbol{F = ma}$

$\longrightarrow \: \sf{F = 7000 \times 2}$

$\therefore \: \boxed{ \bf{Force = 14000 \: N}}$

Answer 2

Given:-

$\mathrm{Initial \: velocity (u) = 0ms}$

$\mathrm{Distance (s) = 400m}$

$\mathrm{Time (t)= 20s}$

• To find acceleration :-

Here, we use 2nd equation of motion :-

$\mathrm{second \: \: eq. \: \: of \: \: motion \: - s = ut + \frac{1}{2} a {t}^{2} }$

So,

$\rightarrow\mathrm{ s = ut + \frac{1}{2} a {t}^{2} }$

$\rightarrow\mathrm{ 400 = 0×20 + \frac{1}{2} ×( a) {(20)}^{2} }$

$\rightarrow\mathrm{ 400= 0 + \cancel\frac{1}{2} (a)× \cancel 400}$

$\rightarrow\mathrm{ 400= 200 \: a}$

$\rightarrow\mathrm{\cancel \frac{400}{200} = a}$

$\rightarrow\mathrm{ a = 2m/ {s}^{2}}$

Therefore,

$\implies\bf \purple{\boxed{\bf {Acceleration = 2m/ {s}^{2} }}}$

● Now, To find force acting on it :-

$\rightarrow\mathrm{Force = mass × acceleration }$

Mass of truck = 7 tonnes

Mass of truck in kg = 7×1000 kg

[1 tonne = 1000kg]

so, mass of truck in kg = 7000 kg

$\rightarrow\mathrm{Force = mass × acceleration }$

$\rightarrow\mathrm{Force = 7000× 2}$

$\rightarrow\mathrm{Force = 14000N }$

Therefore,

$\implies\bf \purple{\boxed{\bf {Force = 14000N }}}$

HOPE IT HELPS YOU DEAR :)