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

Question

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.

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Ladonna 3 years 2021-07-24T05:12:58+00:00 1 Answers 659 views 0

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  1. Sapo1
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    2021-07-24T05:14:37+00:00
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    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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