An object with a mass of 32 kg is attached to a spring ?

An object with a mass of 32 kg is attached to a spring starts from rest and is pulled by the spring across a horizontal frictionless surface and reaches a final velocity of 2.8 m/s in 2s. What is the distance in centimeters that the spring is stretched by if the spring constant is 40 N/m?

a) 11.2

b) 112

c) 2.24

d) 1.12

Please help!

Update:

the answer is 112 but I don't know how to get it 

2 Answers

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  • NCS
    Lv 7
    1 month ago

    Since the surface is frictionless, there is no "final velocity" for the system -- it oscillates forever. The problem cannot be solved as stated.

    Oddly enough, if you solve for the angular frequency, you get one of the choices you list:

    ω = √(k/m) = √(40kg/s² / 32kg) = 1.12 rad/s

    If 2.8 m/s is the MAXIMUM velocity, then the amplitude was

    A = v / ω = 2.8m/s / 1.12rad/s = 2.5 m

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  • Vaman
    Lv 7
    1 month ago

    Let the equation be y= a sin wt.w=sqrt(k/m)= sqrt(40/32)=1.1.   Velocity= aw cos wt. Maximum velocity=2.8/2=1.4= a 1.1, a=distance maximum= 1.4/1.1=1.27 m. d seem to the correct answer.

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