| ⇦ | 
| ⇨ |
A block of mass 10 kg is moving in X-direction with a constant speed of 10 ms⁻¹.It is subjected to a retarding force F=-0.1 x m⁻¹, during its travel from x=20m to x=30m.Its final kinetic energy will be
Options
(a) 475 j
(b) 450 j
(c) 275 j
(d) 250 j
Correct Answer:
475 j
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - In Young’s double slit experiment the distance between the slits and the screen
- If both the resistance and the inductance in an LR AC series circuit are doubled
- Heat is flowing through a rod of length 50 cm and area of cross-section 5 cm².
- A charged particle of mass m and charge q moves along a circular path of radius
- The number of possible natural oscilations of air column in a pipe closed
Topics: Work Energy and Power
(94)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- In Young’s double slit experiment the distance between the slits and the screen
- If both the resistance and the inductance in an LR AC series circuit are doubled
- Heat is flowing through a rod of length 50 cm and area of cross-section 5 cm².
- A charged particle of mass m and charge q moves along a circular path of radius
- The number of possible natural oscilations of air column in a pipe closed
Topics: Work Energy and Power (94)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
a=0.1x/m
=0.1/10=0.01 m/s2
v2=100-0.01[30×30-20×20]
=100-0.01×500
=100-5=95
final kinetic energy =1/2x10x95
=475J