A mass m = 11 kg rests on a frictionless table and accelerated by a spring with

Question

A mass m = 11 kg rests on a frictionless table and accelerated by a spring with spring constant k = 4980 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is 0.46. The mass leaves the spring at a speed v - 2.8 m/s. 1) How much work is done by the spring as it accelerates the mass? 43.12 JSubmit 2) How far was the spring stretched from its unstreched length? 0.132 m Submit Your submissions: 0.1321 Computed value: 0.132 Submitted: Sunday, March 20 at 10:17 PM Feedback: Correct! 3) The mass is measured to leave the rough spot with a final speed vf 1.5 m/s. How much work is done by friction as the mass crosses the rough spot? J Submit

Explanation / Answer

(3)
Work done by Friction, = Change in Kinetic Energy

Initial K.E = 1/2*m*v^2
Final K.E = 1/2*m*vf^2

Work done by Friction, = 1/2*m* (vf^2 - v^2)
Work done by Friction, = 1/2*11*(1.5^2 - 2.8^2)
Work done by Friction, = - 30.75 J
(4)
Let the Length of the spot be x.
Work done by Friction, = Friction Force * Length of the spot
-30.75 = - uk*m*g*x
30.75 = 0.46*11*9.8*x
x = 0.62 m
x = 62.0 cm

(5)
Now, Let the strecth in the string = x
Initial Spring Potential Energy + Work done by Friction = 0
1/2*kx^2 - uk*m*g*(L/2) = 0
1/2*4980*x^2 = 0.46*11*9.8*0.62/2
x =  0.0785 m

(6)
Let the new coefficient be, uk
1/2*kx^2 =
uk*m*g*(L/2) =  uknew*m*g*L
uknew = uk/2
uknew = 0.23

(7)
Work done by Friction, = Friction Force * Length of the spot
As Length is increased three times,

Magnitude of word done will be three times greater.

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