A 500kg block is at the …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Test
Related Questions
Posted by Pankaj Tripathi 1 year, 5 months ago
- 1 answers
Posted by M D 1 year, 4 months ago
- 1 answers
Posted by Nekita Baraily 1 year, 4 months ago
- 2 answers
Posted by Mansi Class 9Th 1 year, 5 months ago
- 0 answers
Posted by Mohammed Javith 1 year, 5 months ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 6 years, 3 months ago
Here,
</div>Mass, m = 500 kg
Inclination of the inclined plane, θ = 30o
Distance traveled along the plane, s = 140 cm = 1.4 m
AB = s = 1.4 m
AC = AB sinθ = (1.4)(sin 30) = (1.4)(1/2) = 0.7 m
Considering C to be the reference level, the potential energy at A is = mg(AC)
This energy is converted into kinetic energy at B = ½ mv2
[since, C is the reference level, the potential energy at C is zero]
So, by energy conservation,
mg(AC) = ½ mv2
{tex}\Rightarrow{/tex} (9.8)(0.7) = ½ v2
{tex}\Rightarrow{/tex}v = 3.7 m/s
This is the velocity at the bottom of the plane.
1Thank You