Question

A particle of mass 0.3 kg is subjected to a force F=-koc
with k = 15 N/m. What will be its initial acceleration if
it is released from a point 20 cm away from the origin



(1) 5 m/s² (2) 10 m/s²

(3) m/s² (4) 15 m/s²​

Answer 1

${ \huge \bf { \mid{ \overline{ \underline{Correct\: Question}}} \mid}}$

A particle of mass 0.3 kg is subjected to a force F= - kx with k = 15 N/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin?

(1) 5 m/s² (2) 10 m/s²

(3) 3 m/s² (4) 15 m/s²

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of the Body (M) = 0.3 Kg.
• Constant (k) = 15 N/m
• Displacement covered (x) = 20 cm.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

From the Given Equation,

$\large{\boxed{\tt F = - kx}}$

Substituting the values,

$\large{\tt \leadsto F = - 15 \times 0.2}$

• As x = 20 cm = 0.2 m (S.I units)

Solving,

$\large{\tt \leadsto F = - 15 \times \dfrac{2}{10}}$

$\large{\tt \leadsto F = \dfrac{- 30}{10}}$

$\large{\tt \leadsto F = \cancel{\dfrac{- 30}{10}}}$

$\large{\boxed{\tt F = - 3 \: N}}$

So, Force Applied is - 3 Newton's.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Newton's second law of motion,

$\large{\boxed{\tt F = Ma}}$

Substituting the values,

$\large{\tt \leadsto - 3 = 0.3 \times a}$

$\large{\tt \leadsto - 3 = \dfrac{3}{10} \times a}$

$\large{\tt \leadsto a = \dfrac{- 3 \times 10}{3}}$

$\large{\tt \leadsto a = \cancel{\dfrac{- 3}{3}} \times 10}$

$\large{\tt \leadsto a = - 10 \: ms^{-2}}$

Taking Magnitude of the Acceleration gives,

$\huge{\boxed{\boxed{\tt a = 10 \: m/s^2 }}}$

So, the Acceleration of the Particle is 10 m/s².

Therefore, Option - (2) is correct.

$\rule{300}{1.5}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

As we have a relation.

$\LARGE \implies {\boxed{\blue{\sf{F \: \propto \: -x}}}}$

Now, we have to remove proportionality

$\Large \implies {\red{\sf{F \: = \: -kx}}}$

Where,

F is Force

x is displacement

k is spring constant

We are given

$\rule{100}{2}$

k = 15 N/m

x = 20 cm => 0.2 m

Now put values in above formula :-

⇒ F = -(15)(0.2)

⇒F = -(3)

⇒F = -3N

$\Large{\boxed{\red{\sf{Force\: (F) \: = \: -3 \: N}}}}$

$\rule{200}{2}$

We also know that

$\LARGE{\boxed{\boxed{\blue{\sf{F \: = \: ma}}}}}$

Now we know,

mass (m) = 0.3 kg

Force (F) = -3 N

Put Values

⇒-3 = (a)(0.3)

⇒-3 = 0.3a

⇒a = -3/0.3

$\Large{\boxed{\red{\sf{Acceleration \: (a) \: = \: -10 \: ms^{-2}}}}}$