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mass hanging mass (kg) force of gravity (n) on hanging mass time (s) di…

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mass hanging mass (kg) force of gravity (n) on hanging mass time (s) displacemen t (m) accelerat (m/s²) 100g (0.1kg) 0.025kg -0.245 250g (0.25kg) 0.025kg -0.245 1.09 0.6 m 1.194 m 500g (0.5kg) 0.025kg -0.245 1.41 0.6 m 750g (0.75kg) 0.025kg -0.245 1.45 0.6 m 1.60 0.6 m calculations 1. force of gravity of hanging mass: .025 × -9.8 = -0.245 2. accelerations: a. 100g 0.6 = 0 + 1/2a(1.0a)² = 1.194 b. 250g c. 500g d. 750g

Explanation:

Step1: Use the kinematic equation

The kinematic equation \(s = ut+\frac{1}{2}at^{2}\). Since the initial velocity \(u = 0\) (starts from rest), the equation simplifies to \(s=\frac{1}{2}at^{2}\). We can solve for \(a\) as \(a=\frac{2s}{t^{2}}\)

Step2: Calculate acceleration for \(100g\) mass

Given \(s = 0.6m\) and \(t = 1.09s\). Substitute into \(a=\frac{2s}{t^{2}}\)

$$ a=\frac{2\times0.6}{(1.09)^{2}}=\frac{1.2}{1.1881}\approx1.01\ m/s^{2} $$

Step3: Calculate acceleration for \(500g\) mass

Given \(s = 0.6m\) and \(t = 1.41s\). Substitute into \(a=\frac{2s}{t^{2}}\)

$$ a=\frac{2\times0.6}{(1.41)^{2}}=\frac{1.2}{1.9881}\approx0.60\ m/s^{2} $$

Step4: Calculate acceleration for \(750g\) mass

Given \(s = 0.6m\) and \(t = 1.45s\). Substitute into \(a=\frac{2s}{t^{2}}\)

$$ a=\frac{2\times0.6}{(1.45)^{2}}=\frac{1.2}{2.1025}\approx0.57\ m/s^{2} $$

Step5: Calculate acceleration for \(1.60s\) time (assuming it's for \(750g\) or other, but using the formula)

Given \(s = 0.6m\) and \(t = 1.60s\). Substitute into \(a=\frac{2s}{t^{2}}\)

$$ a=\frac{2\times0.6}{(1.60)^{2}}=\frac{1.2}{2.56}= 0.47\ m/s^{2} $$

Answer:

  • For \(100g\): \(1.01\ m/s^{2}\)
  • For \(500g\): \(0.60\ m/s^{2}\)
  • For \(750g\) (\(t = 1.45s\)): \(0.57\ m/s^{2}\)
  • For \(t = 1.60s\): \(0.47\ m/s^{2}\)