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3 years ago

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A 200 g block is pressed against a spring of force constant 1.40 kN/m until the block compresses the spring 10.0 cm. The spring

rests at the bottom of a ramp inclined at 60.0 degrees to the horizontal. Using energy considerations, determine how far up the incline the block moves from its initial position before it stops a) if the ramp exerts no friction force on the block and b) if the coefficient of kinetic friction is 0.400.

1 answer:

tino4ka555 [31]3 years ago

6 0

Answer:

Explanation:

This problem bothers on the energy stored in a spring in relation to conservation of energy

Given data

Mass of block m =200g

To kg= 200/1000= 0.2kg

Spring constant k = 1.4kN/m

=1400N/m

Compression x= 10cm

In meter x=10/100 = 0.1m

Using energy considerations or energy conservation principles

The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring

Potential Energy stored in spring

P.E=1/2kx^2

Kinetic energy of the block

K.E =1/mv^2

Where v = velocity of the block

K.E=P.E (energy consideration)

1/2kx^2=1/mv^2

Kx^2= mv^2

Solving for v we have

v^2= (kx^2)/m

v^2= (1400*0.1^2)/0.2

v^2= (14)/0.2

v^2= 70

v= √70

v= 8.36m/s

a. Distance moved if the ramp exerts no force on the block

Is

S= v^2/2gsinθ

Assuming g= 9. 81m/s^2

S= (8.36)^2/2*9.81*sin60

S= 69.88/19.62*0.866

S= 69.88/16.99

S= 4.11m

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