A block of mass m = 2 kg is attached to a spring whose spring constant

1.	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» ⇒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. 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.

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»

⇒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.

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.