Question

A force of 10 N is applied for 0.1
seconds on a body of mass 1 kg
initially at rest. The force then
ceases to act. What would be the
velocity of the body just after this?
After 2 more seconds, what will be
its velocity?

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Velocity\:after\:2\:sec=21\:m/s}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Force(F) = 10 \: N \\ \\ \tt: \implies Mass(M) = 1 \: kg \\\\ \tt:\implies Time(t_{1})=0.1\:sec\\ \\ \tt: \implies Time(t_{2}) = 2 \: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies velocity \: at \: 2 \: sec = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies F = Ma \\ \\ \tt: \implies 10 = 1 \times a \\ \\ \green{\tt: \implies a = 10 { \: m/s}^{2} } \\ \\ \tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \tt \circ \: Time = 0.1 \: sec \\ \\ \tt \circ \: Acceleration = 10 { \: m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies v = 0 + 10 \times 0.1 \\ \\ \tt: \implies v = 0 + 1 \\ \\ \green{\tt: \implies v = 1 { \: m/s}} \\ \\ \tt \circ \: Initial \: velocity = 1 \: m/s \\ \\ \tt \circ \: Time = 2 \: sec \\ \\ \tt \circ \: Acceleration = 10 { \: m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies v = 1 + 10 \times 2 \\ \\ \tt: \implies v =1 + 20 \\ \\ \green{\tt: \implies v = 21 { \: m/s}}\\\\ \green{\tt \therefore Velocity \: after \: 2 \: sec \: is \: 21 \: m/s}$

Answer 2

Aɴꜱᴡᴇʀ

V = 20 m/s

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Gɪᴠᴇɴ

★ F= 10 N

★ Time(1) =0.1 s

★ Time(2) =2 s

★ Mass= 1 kg

★ u = 0m/s

_________________

ᴛᴏ ꜰɪɴᴅ

➜ The velocity after 2 seconds?

_________________

Sᴛᴇᴘꜱ

First let's find the acceleration of the body and as we have the force and the mass of the body we can use the Second law of motion

$\large \sf{ \pink{f = ma}} \\ \\ \red{\sf{} \frac{f}{m} = a} \\ \\ \tt{} \leadsto{} \frac{10}{1} = a \\ \\ \tt{} \dashrightarrow{ \pink{a = 10m/ {s}^{2} }}$

Now let's use the formula v = u+at

If the time was 0.1 seconds

$\tt{} \leadsto{}v = 0 + 10 \times 0.1 \\ \\ \tt{} \dashrightarrow{ \red{v = 1m/s}}$

So if the time was 2 seconds and if we use the same formula we can get our answer,that is

$\tt{} \leadsto{}v = 0 + 10 \times 2 \\ \\ \large { \orange{ \dashrightarrow{ \sf{}v = 20m/s}}}$

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Some related formulas :-

• $\sf{}s = it +\frac{1}{2}a{t}^{2}$

• $\sf {v}^{2} - {u}^{2} = 2as$