FYBSc Internal Assignment Solutions (Explained for Class 11 Level)
📘 FYBSc Internal Assignment Solutions (Explained for Class 11 Level) Q1. A 0.150 kg particle moves along an x-axis according to: x ( t ) = − 13.00 + 2.00 t + 4.00 t 2 − 3.00 t 3 We have to find the net force on the particle at t = 3.40 s t = 3.40 \, \text{s} t = 3.40 s . Step 1: Velocity is derivative of position v ( t ) = d x d t = 2.00 + ( 8.00 t ) − ( 9.00 t 2 ) v(t) = \frac{dx}{dt} = 2.00 + (8.00t) - (9.00t^2) Step 2: Acceleration is derivative of velocity a ( t ) = d v d t = 8.00 − 18.00 t a(t) = \frac{dv}{dt} = 8.00 - 18.00t Step 3: Force using Newton’s 2nd Law F = m ⋅ a At t = 3.40 s t = 3.40 \, \text{s} t = 3.40 s : a ( 3.40 ) = 8.00 − 18.00 ( 3.40 ) = 8.00 − 61.20 = − 53.2 m/s 2 Mass = 0.150 kg F = 0.150 × ( − 53.2 ) = − 7.98 N F = 0.150 \times (-53.2) = -7.98 \, \text{N} 👉 Answer: Net Force = -7.98 N i ^ \hat{i} i ^ (negative means direction is opposite x-axis) Q2. A block of mass 8.5 kg on a 30° incline attached to a cord. We need: (a) Tension in ...