A 10 kg block on a horizontal surface ( μ_k = 0.35 ) is pulled by a 7 kg mass over a pulley. A 40 N force at 37° to th
Given: A 10 kg block on a horizontal surface ( μ_k = 0.35 ) is pulled by a 7 kg mass over a pulley. A 40 N force at 37° to the horizontal opposes the 10 kg block. What is the tension? (Take g = 10 m/s², sin 37° = 0.6, cos 37° = 0.8 ) These values define the system as per NCERT data. Formula: For 7 kg : 7g - T = 7a Rightarrow 70 - T = 7a. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: For 10 kg : T - f_k - F cos 37° = 10a . Normal force: N = mg + F sin 37° = 10 × 10 + 40 × 0.6 = 100 + 24 = 124 N . Friction: f_k = 0.35 × 124 = 43.4 N . F
Ref: NCERT Physics Textbook for Class XI and XII, Chapter: Laws of Motion, Work Energy Power, Gravitation and System of Particles, Topic: Newton's laws, work-energy theorem and rotational dynamics.
