A mass m = 16 kg rests on a frictionless table and accelerated by a spring with

Question

A mass m = 16 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5052 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is k = 0.52. The mass leaves the spring at a speed v = 4 m/s.

1) The mass is measured to leave the rough spot with a final speed vf = 2.3 m/s.How much work is done by friction as the mass crosses the rough spot?

2) What is the length of the rough spot?

3) In this new scenario, what would the coefficient of friction of the rough patch need to be changed to in order for the block to just barely make it through the rough patch?

Explanation / Answer

(1)
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*16*(2.3^2 - 4^2)
Work done by Friction, = - 85.68 J

(2)
Let the Length of the spot be x.
Work done by Friction, = Friction Force * Length of the spot
- 85.68 = - uk*m*g*x
85.68 = 0.52 * 16*9.8*x
x = 1.05 m

(3)
Let the new coefficient be uk.
Now As it barely makes through the patch, Final Veloicty = 0
So,
Work done by Friction, = 1/2*m*(vf^2 - v^2)
- uk*m*g*x = - 1/2*m*4^2
uk*9.8*1.05 = 1/2*16
uk = 0.777
New coefficient of friction, uk = 0.777

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