Block Sliding 2 A mass m = 10 kg rests on a frictionless table and accelerated b

Question

Block Sliding 2 A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k 4461 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is 0.49 The mass leaves the spring at a speed v = 2.8 m/s. How much work is done by the spring as it accelerates the mass? How far was the spring stretched from its unstreched length? The mass is measured to leave the rough spot with a final speed v 1.3 m/s. How much work is done by friction as the mass crosses the rough spot? What is the length of the rough spot? In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length? m Submit 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

Given,

m = 10 kg ; k = 4461 N/m ; uk = 0.49 ; v = 2.8 m/s

1)The work done will be equal to the Kinetic energy of the mass which has been imparetd by the spring.

KE = 1/2 m v^2

KE = 0.5 x 10 x 2.8^2 = 39.2 J

Hence, W = KE = 39.2 J

2)PE(spring) = KE

1/2 k x^2 = 39.2

x = sqrt (2 x 39.2/k) = sqrt (78.4/4461) = 0.133 m

Hence, x = 0.133 m

3)vf = 1.3 m/s

Thw rok done by the frictional surface will be:

Wf = W - KE(f)

Wf = 39.2 - 0.5 x 10 x 1.3^2 = 30.75 J

Hence, Wf = 30.75 J

4)The work done by the patch is:

Wf = uk m g d

d = Wf/ uk m g

d = 30.75/0.49 x 10 x 9.81 = 0.639 m

Hence, d = 0.639 m

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