NEET Laws of Motion — Set 13

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Question 1

A $ 1250 \, \text{kg} $ car turns on a banked road ($ \theta = 15^\circ $, $ \mu_s = 0.3 $) with radius $ 45 \, \text{m} $. What is the maximum speed without slipping? (Take $ g = 10 \, \text{m/s}^2 $, $ \tan 15^\circ \approx 0.268 $)

TopperSquare Expert · Accepted Answer

Maximum speed: $ v_{\text{max}} = \sqrt{rg \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}} $. Numerator: $ \mu_s + \tan\theta = 0.3 + 0.268 = 0.568 $. Denominator: $ 1 - 0.3 \times 0.268 = 1 - 0.0804 = 0.9196 $. $ v_{\text{max}}^2 = 45 \times 10 \times \frac{0.568}{0.9196} \approx 450 \times 0.6175 \approx 277.875 $. $ v_{\text{max}} = \sqrt{277.875} \approx 16.67 \, \text{m/s} $.

Question 2

A $ 0.9 \, \text{kg} $ stone in a vertical circle of radius $ 2 \, \text{m} $ has a tension of $ 30 \, \text{N} $ at the bottom. A tangential force of $ 3 \, \text{N} $ acts upward from bottom to top. What is the tension at the top? (Take $ g = 10 \, \text{m/s}^2 $)

TopperSquare Expert · Accepted Answer

At bottom: $ T_b - mg = m v_b^2 / r \Rightarrow 30 - 0.9 \times 10 = 0.9 \times v_b^2 / 2 $. $ 30 - 9 = 0.45 v_b^2 \Rightarrow 21 = 0.45 v_b^2 \Rightarrow v_b^2 = \frac{21}{0.45} \approx 46.67 \Rightarrow v_b \approx 6.83 \, \text{m/s} $. Work by tangential force: Arc length $ \pi r = \pi \times 2 = 6.28 \, \text{m} $, $ W = F_t \times s = 3 \times 6.28 = 18.84 \, \text{J} $ (reduces KE). Initial KE: $ \frac{1}{2} m v_b^2 = 0.5 \times 0.9 \times 46.67 \approx 21 \, \text{J} $. PE gain: $ 2mg = 2 \times 0.9 \times 10 \times 2 = 36 \, \text{J} $. Final KE: $ 21 - 18.84 - 36 = -33.84 \, \text{J} $ (impossible, adjust: $ v_t^2 \geq 0 $). Correct: $ \frac{1}{2} m v_b^2 - F_t \times 2r = \frac{1}{2} m v_t^2 + 2mg \Rightarrow 21 - 3 \times 4 = 0.45 v_t^2 + 18 $. $ 21 - 12 = 0.45 v_t^2 + 18 \Rightarrow 9 - 18 = 0.45 v_t^2 \Rightarrow v_t^2 < 0 $ (recompute: $ v_t=0 $, $ T_t=0 $, but check). Final: $ v_t^2 = 1.11 $, $ T_t + 9 = 0.5 \Rightarrow T_t = -8.5 $ (impossible, $ T_t = 0 \, \text{N} $ minimum).

Question 3

A $ 0.55 \, \text{kg} $ stone in a vertical circle of radius $ 2.2 \, \text{m} $ has a tension of $ 30 \, \text{N} $ at the bottom. What is the tension at the top? (Take $ g = 10 \, \text{m/s}^2 $)

TopperSquare Expert · Accepted Answer

At bottom: $ T_b - mg = m v_b^2 / r \Rightarrow 30 - 0.55 \times 10 = 0.55 \times v_b^2 / 2.2 $. $ 30 - 5.5 = 0.25 v_b^2 \Rightarrow 24.5 = 0.25 v_b^2 \Rightarrow v_b^2 = 98 \Rightarrow v_b \approx 9.9 \, \text{m/s} $. Energy: $ \frac{1}{2} m v_b^2 = \frac{1}{2} m v_t^2 + 2mg $. $ 0.5 \times 0.55 \times 98 = 0.5 \times 0.55 \times v_t^2 + 2 \times 0.55 \times 10 $. $ 26.95 = 0.275 v_t^2 + 11 \Rightarrow 0.275 v_t^2 = 15.95 \Rightarrow v_t^2 \approx 58 \Rightarrow v_t \approx 7.62 \, \text{m/s} $. At top: $ T_t + mg = m v_t^2 / r \Rightarrow T_t + 5.5 = 0.55 \times 58 / 2.2 \Rightarrow T_t + 5.5 \approx 14.5 \Rightarrow T_t \approx 9 \, \text{N} $.

NEET Physics — Laws of Motion

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Q1. A \( 1250 \, \text{kg} \) car turns on a banked road (\( \theta = 15^\circ \), \( \mu_s = 0.3 \)) with radius \( 45 \, \text{m} \). What is the maximum speed without slipping? (Take \( g = 10 \, \text{m/s}^2 \), \( \tan 15^\circ \approx 0.268 \))

Q2. A \( 0.9 \, \text{kg} \) stone in a vertical circle of radius \( 2 \, \text{m} \) has a tension of \( 30 \, \text{N} \) at the bottom. A tangential force of \( 3 \, \text{N} \) acts upward from bottom to top. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

Q3. A \( 0.55 \, \text{kg} \) stone in a vertical circle of radius \( 2.2 \, \text{m} \) has a tension of \( 30 \, \text{N} \) at the bottom. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

Q4. A \( 6.2 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.25 \)) is connected to a \( 8.2 \, \text{kg} \) mass over a pulley. A \( 15 \, \text{N} \) force aids the \( 6.2 \, \text{kg} \) block. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

Q5. A \( 5.6 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.2 \)) is pulled by a \( 3.6 \, \text{kg} \) mass over a pulley. A \( 11 \, \text{N} \) force opposes the \( 5.6 \, \text{kg} \) block. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

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NEET Physics: Laws of Motion — MCQ Test 13

Q6. A \( 4.5 \, \text{kg} \) block on a \( 60^\circ \) incline (\( \mu_s = 0.35 \)) is pulled downward by a horizontal force just sufficient to start motion. What is the force? (Take \( g = 10 \, \text{m/s}^2 \), \( \sin 60^\circ = 0.866 \), \( \cos 60^\circ = 0.5 \))

Q7. A \( 0.65 \, \text{kg} \) stone in a vertical circle of radius \( 1.6 \, \text{m} \) has a tension of \( 35 \, \text{N} \) at the bottom. What is the tension at the top? (Take \( g = 10 \, \text{m/s}^2 \))

Q8. A \( 4.2 \, \text{kg} \) block on a horizontal surface (\( \mu_k = 0.2 \)) is connected to a \( 6.2 \, \text{kg} \) mass over a pulley. A \( 16 \, \text{N} \) force aids the \( 4.2 \, \text{kg} \) block. What is the tension? (Take \( g = 10 \, \text{m/s}^2 \))

Q9. A \( 0.28 \, \text{kg} \) stone is whirled in a horizontal circle of radius \( 1.4 \, \text{m} \) with a tension of \( 12 \, \text{N} \). What is the speed? (Take \( g = 10 \, \text{m/s}^2 \))

Q10. A \( 2100 \, \text{kg} \) rocket accelerates upward at \( 2 \, \text{m/s}^2 \) with a thrust of \( 27000 \, \text{N} \). What is the mass ejection rate if the gas speed is \( 75 \, \text{m/s} \) relative to the rocket? (Take \( g = 10 \, \text{m/s}^2 \))