Question

A body of mass 500 g is at rest on a frictionless surface. what is the distance travelled by it in 10s when acted upon by a force of 10^-2 N? ​

Answer 1

$\bf\pink{Answer}$

$\sf\red{Given - }$

$\bf m = 500g = 0.5 kg$

$\bf t = 10 sec$

$\bf F = 10^{-2}$

$\bf u = 0$

where

$\longrightarrow$m is mass of body.

$\longrightarrow$u is initial velocity.

$\longrightarrow$t is time taken.

$\longrightarrow$F is force applied.

━━━━━━━━━━━━━

$\sf\red{To\: find -}$

Distance travelled $\longrightarrow$ s

━━━━━━━━━━━━━

$\sf\red{Formula \: used -}$

$\boxed{\bf \pink {F = ma}}$

$\boxed{\bf \pink {s = ut + 1/2 at^2}}$

where

$\longrightarrow$m is mass of body.

$\longrightarrow$t is time taken.

$\longrightarrow$F is force applied.

$\longrightarrow$u is initial velocity.

$\longrightarrow$a is acceleration.

━━━━━━━━━━━━━

$\sf\red{Solution - }$

$\longrightarrow$$F = 10^-2$

$\longrightarrow$$m = 500g = 0.5 kg$

Using the formula -

$\bf \pink {F = ma}$

$\implies$$\bf \green {10 ^{ - 2} = 0.5 \times a }$

$\implies$$\bf \green {a = 0.01 / 0.5 }$

$\implies$$\bf \green {a = 0.02 m/s^2 }$

━━━━━━━━━━━━━

$\longrightarrow$$\bf t = 10 sec$

$\longrightarrow$$\bf a = 0.02 m/s^2$

$\longrightarrow$$\bf u = 0 m/s$

By using the second equation of motion -

$\bf \pink {s = ut + 1/2 at^2}$

$\implies$$\bf \green {s = \frac{1}{2} \times 0.02 \times 10^{2} }$

$\implies$$\bf \blue {s = 1m}$

Distance travelled by body in 10 seconds is 1 m

━━━━━━━━━━━━━

$\bf\pink{Thanks}$

Answer 2

$\underline {\underline{{\red{ \sf Given }}}}$

• Mass (m) ➠ 500g ➠ 0.5kg
• Time (t) ➠ 10 s
• Force Applied (N) ➠ 10⁻²
• Initial Speed (u) ➠ 0

$\underline {\underline{{ \red{\sf To \: Find }}}}$

• Distance Travelled (s)

Firstly,we will calculate a Acceleration and finally we will calculate the distance travelled

$\huge{ \underline {\underline{{ \green{\sf Calculating \: Acceleration }}}}}$

Formula Used :- $\underline{\underline{\boxed{ \gray{\sf Force (F) = Mass(m) \times Acceleration (a) }}}}$

Substituting Values

☞ 10⁻² = 0.5 × a

☞ 10⁻² = 0.5a

☞ 0.5a = 0.01

☞ a = ⁰∙¹⁄₀.₀₅

☞ a = 0.02

${\pink{ \underbrace{\boxed{\underline{\underline{\blue{ \tt Thus, we \: got \: acceleration \: 0.02 \: m/s^{2}}}}}}}}$

Now,

$\huge{ \underline {\underline{{ \green{\sf Calculating \: Distance \: Travelled }}}}}$

Formula Used :- $\underline{\underline{\boxed{ \gray {\sf S = ut + \dfrac{1}{2} at^{2}}}}}$

Substituting Values

☞ s = 0 × 10 + ½ × 0.02 × (10)²

☞ s = 0 + ½ × 0.02 × 100

☞ s = ½ × 2

☞ s = ²⁄₂

☞ s = 1

${\red{ \underbrace{\boxed{\underline{\underline{\blue{ \tt Therefore, Distance \: Travelled \: is \: 1 \: m}}}}}}}$