Question

A $5.0$kg mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in the FIGURE. The scale reads in newtons.

a. What does the spring scale read just before the mass touches the lower spring?

b. The scale reads $20$N when the lower spring has been compressed by $2.0$cm. What is the value of the spring constant for the lower spring?

c. At what compression will the scale read zero?

Short answer

a). The spring scale reads $49$N

b). The spring constant is $k=1450$N/m

c). Compression will the scale read zero $∆s=3.3$cm

Step-by-step solution

Step 1. Given Information

A mass hanging $5.0$kg

The exerts force$20N$

Step 2. Part a). Before the mass touches the lower spring

The spring scale reads only the weight of the mass which is given by

$w=mg$

Put the values

$$\begin{gathered} w=mg \\ =\left(5kg\right)(9.8m/s^2) \\ =49N \end{gathered}$$

Step 3. Part b). The spring constant for the lower spring

The mass exert force $49$N and the scale reads $20$N, this means the difference between these two values represents the restoring force of the spring

$F_{sp}=49N-20N=29N$

Hooke's law state the relationship between the restoring force and the displacement $∆s$

$F_{sp}=k∆s$

Now,

$$\begin{gathered} k=\frac{F_{sp}}{∆s} \\ =\frac{29N}{0.02m} \\ =1450N/m \end{gathered}$$

Step 4. Part c). Finding compression will the scale read zero.

The scale reads zero when the restoring force equals the weight of the mass which is $49$N.

So,

$$\begin{gathered} ∆s=\frac{F_{sp}}{k} \\ =\frac{49N}{1450N/m} \\ =0.033m \\ =3.3cm \end{gathered}$$