A mass m = 12 kg rests on a frictionless table and accelerated by a spring with

Question

A mass m = 12 kg rests on a frictionless table and accelerated by a spring with spring constant k = 4353 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is k = 0.43. The mass leaves the spring at a speed v = 2.9 m/s.

1)

How much work is done by the spring as it accelerates the mass? J

ANS: 50.46 J (CORRECT; I did .5mv^2)

2)

How far was the spring stretched from its unstreched length? m

3)

The mass is measured to leave the rough spot with a final speed vf = 1.7 m/s.

How much work is done by friction as the mass crosses the rough spot? J

4)

What is the length of the rough spot? m

5)

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

6)

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?

7)

Return to a scenario where the blcok makes it throgh the entire rough patch. If the rough patch is lengthened to a distance of three times longer, as the block slides through the entire distance of the rough patch, the magnitude of the work done by the force of friction is:

the same

three times greater

three times less

nine times greater

nine times less

(SOLUTIONS WITH FORMULAS WILL BE GREATLY APPRECIATED!)

Explanation / Answer

  1)

W = (1/2)mv^2 = (0.5) (12) (2.9)^2 = 50.46 J

2)

W = (1/2)kx^2

50.46 = (0.5) (4353) x^2

x = 0.1523 m

3)

KE1 = KE2 + frictionWork

FW = KE1 - KE2

= 50.46 - 1/2*m*v2^2

= 50.46 - 1/2*12*1.7^2

= 50.46 - 17.34

= 33.12 J

4)

F = uk*N = uk*m*g

W = F*d = (uk*m*g) * d

d = W/F = W/(uk*m*g)

= 33.12/(0.43*12*9.81)

= 0.654 m

5)

d = 0.327 m

W = uk*m*g*d

W = 0.43*12*9.81*0.327

W = 16.55 J

SE = W

1/2*k*x^2 = W

x = sqrt(2*W/k)

x = sqrt(2*16.55/4353)

x = 0.0872 m

6)

W = d*uk*m*g

let d = half of rough patch

change uk so that mass barely travels through rough patch instead of halfway

W =d*uk*m*g = (2d)*(x*uk)*m*g

x = 1/2

so reduce uk by half

new uk = 0.215

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.