Consider a block and trolley system as shown in figure. If the coefficient of kinetic friction between the trolley and the surface is 0.04, the acceleration of the system in mathrmms^-2 is (Consider that the string is massless and unstretchable and the pulley is also massless and frictionless):
Block and trolley tension system for Q48 - JEE Main 2024 Morning
A block and trolley mass arrangement demonstrating a horizontal kinetic interface connected over a corner pulley driven by an explicit 60N forcing function loop.

Solution & Explanation

### Related Formula Kinetic friction force: f_k = mu_k N = mu_k m_1 g System acceleration: a = fracF_textpull - f_km_texttotal ### Core Logic Given values: Trolley mass m_1 = 20mathrm~kg, total system mass component in frame m_texttotal = 26mathrm~kg (from solution fraction frac60-826). Applied pulling force F = 60mathrm~N. mu_k = 0.04. Calculate the kinetic friction resisting the trolley: f_k = 0.04 times 20 times 10 = 8mathrm~N ### Step 1: Calculate Acceleration Using the dynamic equation for connected translation systems: a = frac60 - 826 = frac5226 = 2mathrm~ms^-2 ### Pattern Recognition Treat connected inline systems as a single collective mass block, balancing external driving forces against collective internal friction resistance. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion

Reference Study Guides

More Laws of Motion Previous-Year Questions — Page 6

Q47 jee_main_2024_31_jan_morning Circular Motion Friction
A coin is placed on a disc. The coefficient of friction between the coin and the disc is mu. If the distance of the coin from the center of the disc is r, the maximum angular velocity which can be given to the disc, so that the coin does not slip away, is:
  • A. fracmu gr
  • B. sqrtfracrmu g
  • C. sqrtfracmu gr
  • D. fracmusqrtrg

Solution

### Related Formula f_s leq mu_s N F_c = mromega^2 ### Core Logic
Circular Motion Friction diagram for Q47 - JEE Main 2024 Morning
Circular Motion Friction diagram for Q47 - JEE Main 2024 Morning
Circular Motion Friction diagram for Q47 - JEE Main 2024 Morning
Circular Motion Friction diagram for Q47 - JEE Main 2024 Morning
To prevent the coin from slipping, the static friction must provide the necessary centripetal force for circular motion. f = momega^2 r The normal force on the flat disc is N = mg. The maximum static friction is f_textmax = mu N = mu mg. For no slipping: m r omega^2 leq mu mg omega^2 leq fracmu gr omega_textmax = sqrtfracmu gr ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws Of Motion

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