Question

a body of mass 100 g is at rest on a surface a force of o.2 Newton acts on it for 5 second calculate the distance travelled by the body​

Answer 1

Anѕwєr :

25 m

Exρlαnαtion :

As per the provided information in the given question, we have :

• Mass of the body, m = 100g = 0.1 kg
• Initial velocity, u = 0 m/s [At rest]
• Force acting, F = 0.2 N
• Time taken, t = 5 seconds

We've been asked to calculate the distance travelled by the body.

In order to calculate the distance travelled by the body, firstly we need to calculate the acceleration and in order to calculate the acceleration, we need to find out the final velocity first.

We'll be using the formula of Force derived using the 2nd law of motion to calculate the final velocity. We know that,

$\bigstar \; \underline{\boxed{\bf{F = ma}}}$

• F denotes force
• m denotes mass
• a denotes acceleration

$\implies \rm{ F = m\Bigg ( \dfrac{v-u}{t} \Bigg) \; \; \; \because \: \Bigg \{ a = \dfrac{v-u}{t} \Bigg \} }$

$\implies \rm{0.2 = 0.1 \Bigg(\dfrac{v-0}{5} \Bigg ) }$

$\implies \rm{\dfrac{0.2}{0.1} = \Bigg(\dfrac{v}{5} \Bigg ) }$

$\implies \rm{ \cancel{\dfrac{20}{10} }= \dfrac{v}{5} }$

$\implies \rm{ 2 = \dfrac{v}{5} }$

$\implies \rm{ 2 \times 5 = v }$

$\implies \boxed{\rm{ 10 \; ms^{-1} = v }}$

Now, calculating the acceleration. As we know that,

$\bigstar \; \underline{\boxed{\bf{a = \dfrac{v-u}{t} }}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

$\implies \rm{ a = \dfrac{10 - 0}{5} }$

$\implies \rm{ a = \cancel{\dfrac{10}{5} }}$

$\implies \boxed{\rm{a = 2 \; ms^{-2} }}$

Now, by using the 2nd equation of motion,

$\bigstar \; \underline{\boxed{\bf{s = ut + \dfrac{1}{2}at^2 }}}$

• s denotes distance
• u denotes initial velocity
• t denotes time
• a denotes acceleration

$\implies \rm{ s = 0(5) + \dfrac{1}{2} \times 2\times (5)^2 }$

$\implies \rm{ s = \dfrac{1}{1} \times 1\times 25 }$

$\implies \; \underline{\boxed{\bf{s = 25 \; m }}}$

∴ The distance travelled by the body is 25 m.

$\rule{200}2$