Question

In the figure(Figure 1) a 1.5 kg block is held at rest against a spring with a force constant k = 720 N/m . Initially, the spring is compressed a distance d. When the block is released, it slides across a surface that is frictionless except for a rough patch of width 5.0 cm that has a coefficient of kinetic friction μk = 0.45. Part A Find d such that the block's speed after crossing the rough patch is 2.4 m/s . Express your answer using two significant figures.

Answer 1

Using Energy conservation:

KEi + PEi + Wf = KEf + PEf

KEi = 0, since initially block is at rest

KEf = (1/2)*m*Vf^2

Vf = final speed of block = 2.4 m/sec

PEi = (1/2)*k*d^2

d = initial compression of spring = ?

Wf = work-done by friction force = -Ff*x

Ff = friction force = -uk*N = -uk*m*g*x

x = horizontal rough patch = 5.0 cm = 0.05 m

PEf = 0, since final compression in spring = 0

So,

0 + (1/2)*k*d^2 - uk*m*g*x = (1/2)*m*Vf^2 + 0

k*d^2 = m*Vf^2 + 2*uk*m*g*x

d = sqrt [m*Vf^2/k + 2*uk*m*g*x/k]

m = mass of block = 1.5 kg

k = spring constant = 720 N/m

uk = coefficient of kinetic friction = 0.45

Using these values:

d = sqrt (1.5*2.4^2/720 + 2*0.45*1.5*9.81*0.05/720)

d = 0.1137 m

In two significant figures

d = 0.11 m = initial compression in spring

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Answer 2

Solution:

1. Energy Conservation:
   The block's initial spring potential energy converts to kinetic energy minus work done by friction.

2. Equations:

• Spring energy: $\frac{1}{2}k{d}^2$

• Kinetic energy after rough patch: $\frac{1}{2}m{v}^2$ (where $v=2.4\text{m/s}$)

• Work done by friction: ${\mu }_{k}mg\times \text{width}$

3. Solve for $d$:

   $$\frac{1}{2}k{d}^2=\frac{1}{2}m{v}^2+{\mu }_{k}mg\times 0.05\text{m}$$

   Plug in values ($m=1.5\text{kg},k=720\text{N/m},{\mu }_k=0.45$):

   $$d=\sqrt{\frac{m{v}^2+2{\mu }_{k}mg\times 0.05}{k}}\approx 0.12\text{m}$$

Answer: The spring compression distance $d$ is 0.12 m (12 cm).