Question

A particle of mass m = 0.2 kg has an initial speed of 5 ms-¹ at the bottom of a rough inclined plane of inclination 30° and vertical height 0.5m. Find the speed of the particle as it reaches the top of the inclined plane. (union= 1/√3,g = 10 ms-²)​

Answer 1

Explanation:

Answer

Given,

Masses=5kg,2kg

So,

Let x be the extension when t kg leaves the contact.

kx=2g

x=

k

2g

=

2

1

m

So, by the law of conservation of energy for 5kg mass

mgx=

2

1

(x

2

+mv

2

)

=

2

1

x

2

+

2

mv

2

⇒

2

2mgx−x

2

=

2

mv

2

v=

2gx−

m

kx

2

=

2×

2

101

−

5

40

×

4

1

=2

2

m

THus, the value of x is =2

2

m