pls answer with a diagram with proper explanation

Question

A block of mass m = 2 kg is attached to a spring whose spring constant is k = 8 Nm^-1. The block slides on an incline for and \(\theta\) = 37°. If the block starts at rest with spring un extended, what is its speed, in ms^-1, when it has moved a distance d = 0.5 m down the incline?

Answer 1

⇒Contact force is the resultant of Frictional force and normal reaction force. 
The frictional force can be calculated as

f=0.5xmgcosθ.

f=0.5x2x10x\(\sqrt\frac{3}{2}\)​​

⇒normal reaction force is

 mgcosθ.=2x10×\(\sqrt\frac{3}{2}\)

They are at 90 degree to each other therefore find out their resultant and that shall be the resultant contact force.

Answer 2

Contact force is the resultant of Frictional force and normal reaction force. 

The frictional force can be calculated as =0.5xmgcosθ.=0.5x2x10x (3/2)^1/2

normal reaction force is mgcosθ.=2x10×(3/2)^1/2

They are at 90 degree to each other therefore find out their resultant and that shall be the resultant contact force.