KCET · Physics · Mechanical Properties of Fluids
A \( 10 \mathrm{~kg} \) metal block is attached to a spring of spring constant \( 1000 \mathrm{~N} \mathrm{~m}^{-1} \), A block is
displaced from equilibrium position by \( 10 \mathrm{~cm} \) and released. The maximum acceleration of the
block is
- A \( 10 \mathrm{~ms}^{-2} \)
- B \( 100 \mathrm{~ms}^{-2} \)
- C \( 200 \mathrm{~ms}^{-2} \)
- D \( 0.1 \mathrm{~ms}^{-2} \)
Answer & Solution
Correct Answer
(A) \( 10 \mathrm{~ms}^{-2} \)
Step-by-step Solution
Detailed explanation
Given, mass, \(\mathrm{m}=10 \mathrm{~kg}\); spring constant \(k=1000 \mathrm{Nm}^{-1}\), displacement \(x=10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}\)
We know \(\mathrm{F}=\mathrm{kx}\)
Also, \(\mathrm{F}=\mathrm{ma}\)
\(\Rightarrow \mathrm{kx}=\mathrm{ma}\)
\(\Rightarrow a=\frac{k x}{\mathrm{~m}}=\frac{1000 \times 10 \times 10^{-2}}{10}=10 \mathrm{~ms}^{-2}\)
Therefore, maximum acceleration of the block is \(10 \mathrm{~ms}^{-2}\)
We know \(\mathrm{F}=\mathrm{kx}\)
Also, \(\mathrm{F}=\mathrm{ma}\)
\(\Rightarrow \mathrm{kx}=\mathrm{ma}\)
\(\Rightarrow a=\frac{k x}{\mathrm{~m}}=\frac{1000 \times 10 \times 10^{-2}}{10}=10 \mathrm{~ms}^{-2}\)
Therefore, maximum acceleration of the block is \(10 \mathrm{~ms}^{-2}\)
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