Question

$\mathscr\red{HOLA}$

Questions of the day..


Solve both questions...


1. A force of 10 N acts on a particle of mass 0.4 kg. Find the acceleration of the particle. Change units also.

2. A body of mass 1 Kg is kept at rest. A constant force of 6.0 N starts acting on it.Findb the time taken by the body to move through a distance of 12 m .


Also the answers are 0.04 N and 25m/s^2.


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Full solution.

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Answer 1

$\large{\mathfrak{\red{\underline{\underline{Question-1}}}}}$

A force of 10 N acts on a particle of mass 0.4 kg. Find the acceleration of the particle.

$\large{\mathfrak{\red{\underline{\underline{Solution-1}}}}}$

$\tt{Given:-}$

$\sf{Force=10\;N}$

$\sf{Mass=0.4\;kg}$

$\sf{Acceleration=??}$

$\sf{By\;using\;F=ma}$

$\sf{F = ma}$

$\sf{10 = 0.4\;a}$

$\sf{a=\dfrac{10}{0.4}}$

${\boxed{\boxed{\bf{a = 25\;m/s^{2}}}}}$

--------------------------------------------------------------------------------------------------------------

$\large{\mathfrak{\red{\underline{\underline{Question-2}}}}}$

A body of mass 1 Kg is kept at rest. A constant force of 6.0 N starts acting on it. Find the time taken by the body to move through a distance of 12 m.

$\large{\mathfrak{\red{\underline{\underline{Solution-2}}}}}$

$\tt{Given:-}$

$\sf{F=6\;N}$

$\sf{m=1\;kg}$

$\sf{u=0\;m/s}$

$\sf{s=12\;m}$

$\sf{By\;using\;F=ma,firstly\;we\;find\;a}$

$\sf{F=ma}$

$\sf{6\;N=1\;kg \times a}$

$\bf{\implies a = 6\;m/s^{2}}$

$\sf{By\;using\;2nd\;equation\;of\;motion,}$

$\sf{s=ut+\dfrac{1}{2}\;at^{2}}$

$\sf{12 = 0+\dfrac{1}{2}\times 6 \times t^{2}}$

$\sf{12=3t^{2}}$

$\sf{t^{2}=4}$

${\boxed{\boxed{\bf{t=2\;sec}}}}$

Answer 2

Answer:

Hi Alexa, seen this question now...

1. F=ma

10N=0.4 kg x a

10/0.4=a

a=25m/s sqaure

2. F=ma

6N= 1kg x a

a=6m/s sqaure

FROM 2ND EQAUTION OF MOTION..

s=ut + 1/2at sqaure

12=0+1/2x6 x t square

12=3t sqaure

t= 2 sec...

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