A body of mass 2mathrm~kg moving with velocity of vecv_mathrmin = 3hati +4hatjmathrm~m/s enters into a constant force field of 6mathrm~N directed along positive z-axis. If the body remains in the field for a period of frac53 seconds, then velocity of the body when it emerges from force field is:

Solution & Explanation

### Related Formula vecF = m veca vecv = vecu + vecat where, vecF = constant force vector m = mass of body veca = acceleration vector vecu = initial velocity vector vecv = final velocity vector ### Core Logic Given parameters: - Mass, m = 2mathrm~kg - Initial velocity, vecu = 3hati + 4hatjmathrm~m/s - Force, vecF = 6hatkmathrm~N (directed along positive z-axis) - Time interval, t = frac53mathrm~s Calculate the acceleration vector veca: veca = fracvecFm = frac6hatk2 = 3hatkmathrm~m/s^2 ### Step 1: Compute Final Velocity Using the kinematic equation of motion: vecv = vecu + vecat vecv = (3hati + 4hatj) + (3hatk) left(frac53right) vecv = 3hati + 4hatj + 5hatkmathrm~m/s Thus, the emerging velocity of the body is 3hati + 4hatj + 5hatkmathrm~m/s. ### Pattern Recognition Sees: Orthogonal initial velocity and force field direction. Shortcut: Since the force acts entirely along the z-axis, the x and y components of the velocity remain unchanged (3hati + 4hatj). Simply compute the z-component change: v_z = a_z t = left(frac62right) left(frac53right) = 5. Result: 3hati + 4hatj + 5hatk. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Laws of Motion Class 11 Physics: Kinematics

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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