Question

A 7.8kg mass hanging from a spring scale is slowly lowered onto a vertical spring, as shown in the figure.
What does the spring scale read just before the mass touches the lower spring?(N)
The scale reads 22N when the lower spring has been compressed by 2.6cm . What is the value of the spring constant for the lower spring? (N/m)

At what compression length will the scale read zero? (cm)

Answer 1

Concepts and reason

The concepts used to solve this problem are Hooke’s law of the motion of a mass which is hung on spring balance that compresses the spring, and net force acting on the lower spring.

First, use the force affecting the spring scale to calculate the spring scale reading, which touches the lower spring.

Then, use the spring scale reading and the new scale reading to find the force of lower spring that pushes upward.

Then apply Hooke’s law to calculate spring constant of the spring compressed by 2.0 cm.

Finally, use the spring constant for the lower spring to calculate the compressed length that will read zero in spring scale.

Fundamentals

When the mass rests on the lower spring, the force acting on the mass is weight.

The expression for the weight of the mass is,

Here, W is the weight of the mass, m is the mass hanging in the spring scale, and g is the acceleration due to gravity.

Hooke’s law states that the stretch of the spring from its rest position is proportional to the applied force.

Express the relation using Hooke’s law.

$F=-kor $

Here, F is the applied force, k is the spring constant, and x is the compressed length.

(a)

The only force affecting the spring scale before the mass touches the vertical spring is the weight of the mass.

Hence, the only force acting on the spring balance is weight of the mass.

Substitute for m and for g.

(b)

When the mass is hanging from the spring scale, it reads. When the mass from a spring scale is lowered on to a lower vertical spring, the spring balance reading decreases. This is because the restoring force from the vertical spring pushes the mass from the scale upward.

In equilibrium, the acceleration is zero, so the net force is zero. Thus, the spring force and force due to gravity must balance each other.

Since the spring scale reads , the force of lower spring that pushes upward is,

$W=F+W $

Here, is the new scale reading and is the restoring force from the lower spring.

is acting in the direction opposite to the weight of the mass hanging from the scale. Thus,

Substitute for.

$W=(-kat)+W $

Rearrange the equation for spring constant,

Substitute for , for , and for .

(c)

The scale reads zero when the weight of the hanging mass is completely balanced by the restoring force of the lower vertical spring.

The equation for compression length at which the scale read zero is,

Substitute for .

Rearrange the equation to compressed length at which the spring should be zero,

Substitute for and for .

Ans: Part a

Spring scale reads just before the mass touches the lower spring is .

Part b

The spring constant for the lower spring is .

Part c

Compressed length at which the spring should be zero is .