A mass m = 13 kg rests on a frictionless table and accelerated by a spring with

Question

A mass m = 13 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5059 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is ?k = 0.44. The mass leaves the spring at a speed v = 3.7 m/s.

1)

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

2.

How far was the spring stretched from its unstreched length?

3.

What is the length of the rough spot?

4.

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?

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?

Explanation / Answer

Part 1)

Work done = KE = 1/2mv^2 = ½*13*3.7^2 = 88.985 J

Part 2)

Now let us calculate the ‘x’ made by the spring

Use equation,

KE=1/2mv^2=1/2k x^2

x=sqrt(m/k) = v*sqrt(m/k) = 3.7*sqrt(13/5059) = 0.1876m

Part 3)

Incomplete information for this part. Velocity of mass after leaving the patch is needed.

or

It must be assume that mass stops at the end of the patch.

If it stops at the end, assume length of the patch = d

KE= frictional work done at rough patch

uk*Fn*d = 1/2mv^2

uk*mg*d = 1/2mv^2

0.44*13*9.8*d = 88.985       => d= 1.6m

part 4)

When mass make exactly half-way d=1.6/2= 0.8m

KE= uk*mg*d =1/2mv^2=1/2k x^2

uk*mg*d=1/2k x^2

0.44*13*9.8*0.8=1/2*5059* x^2                     => x = 0.133m

Part 6)

Incomplete information again here

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