⇦ | ![]() | ⇨ |
A block of mass 5 kg is moving horizontally at a speed of 1.5 m/s. A perpendicular force of 5 N acts on it for 4 sec. What will be the distance of the block from the point where the force started acting?
Options
(a) 6 m
(b) 8 m
(c) 10 m
(d) 2 m
Correct Answer:
10 m
Explanation:
Horizontal distance,x=ut ⇒1.5 x 4 = 6m
Vertical distance,y=(1/2)at² = (1/2)(F/m)t²
Two motions are in mutually perpendicular directions. Net displacement = √(6²+8²) = √(36+64) =√100 = 10m.
Related Questions:
- A thin film of soap solution (n=1.4) lies on the top of a glass plate (n=1.5).
- An electric dipole of moment p is lying along a uniform electric field E. The work
- A bus is moving with a velocity of 5 ms⁻¹ towards a huge wall.
- A string is stretched between fixed points seperated by 75 cm. It is observed
- Light of frequency 8×10¹⁵ Hz is incident on a substance of photoelectric work
Topics: Laws of Motion
(103)
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
Leave a Reply