Physics Help? 2

Question

A spring of force constant k = 5000 N/m is placed at the bottom of a frictionless inclined plane that makes 35.0° with the horizontal. The spring is compressed 1.20 cm; then a block of mass 45.0 g is placed at its upper end. How far from its starting position (from the compressed spring) does the block move up the inclined plane?

 b.  Suppose that the coefficient of kinetic friction between the block and the inclined plane is 0.300.   How far from its starting position does the block move up the plane in this case?

Answer 1

(a)

Step 1:

Spring constant k = 5000 N/m.

Angle of inclination is 35.0°.

Spring is compressed by a distance 1.20 cm = 0.012 m

Weight of the block of the mass is 45.0 g = 0.045 kg.

When the block hits the upper end of the spring, spring is compressed by a distance of x.

Then the block moves upward to a distance of d.

So the distance travelled by the block is x +d .

Potential energy of the spring .

Kinectic energy of the block .           (If force of friction is neglected)

According to the law of conservation of energy : .

Block moves up by a distance .

Solution :

Block moves up by a distance .

Answer 2

(b)

Step 1:

Spring constant k = 5000 N/m.

Angle of inclination is 35.0°.

Spring is compressed by a distance 1.20 cm = 0.012 m.

Weight of the block of the mass is 45.0 g = 0.045 kg.

Coefficient of friction is 0.300.

When the block hits the upper end of the spring, the spring is compressed by a distance of x and the block moves upward to a distance of d.

So the distance travelled by the block is x +d .

Potential energy of the spring .

Kinectic energy of the block .

According to the law of conservation of energy : .

Block moves up by a distance .

Solution :

Block moves up by a distance .