Question

A man is slipping on a frictionless inclined plane & a bag falls down from the same height. Then the speed of both is related as:

• A. \(V_B >V_m\)
• B. \(V_B<V_m\)
• C. \(V_B=V_m\)
• D. \(V_B\  and \ V_m\  can't\  related\)

Answer 1

The Correct Option is C

• The problem involves two objects: a man slipping down a frictionless inclined plane and a bag falling freely from the same height.
• We need to determine the relationship between the speeds of the man (\(V_m\)) and the bag (\(V_B\)) when they both reach the ground.

Step-by-Step Solution:

1. Concept Understanding: In physics, when an object falls freely under gravity or slides down a frictionless inclined plane, it primarily undergoes gravitational acceleration.
2. Energy Consideration for the Bag:
   • The bag falls vertically under the influence of gravity, converting its potential energy entirely into kinetic energy as it reaches the ground.
   • Using the energy conservation principle: mgh = \frac{1}{2} mV_B^2, where h is the initial height.
   • Simplifying gives: V_B = \sqrt{2gh}
3. Energy Consideration for the Man:
   • The man slips down a frictionless incline, also converting potential energy to kinetic energy.
   • Assuming the incline angle allows the same vertical drop, the energy equation is the same: mgh = \frac{1}{2} mV_m^2
   • Simplifying: V_m = \sqrt{2gh}
4. Conclusion: From the equations derived:
   • V_B = V_m = \sqrt{2gh}
   • Both the man and the bag reach the same speed upon hitting the ground, since they both convert their potential energy (from the same height) entirely into kinetic energy, driven solely by gravity.

Thus, the correct answer is: \(V_B=V_m\).