Question

The length of a spring becomes 10 cm on
suspending a mass of 20 kg in a vertical plane
and 12 cms on suspending 32 kg. What should
be the weight suspended from it so as to cause
the length to be 15 cms (g = 10 m/sec):
(1) 40 kg
(2) 50 kg
(3) 60 kg
(4) 80 kg

Answer 1

Answer:

50 kg it is used calibration of weighing machines take the constant term

as the stretchiness is directly proportional to the weight tied to it

therefore the the answer is 50 kg by using Newton 2nd law of motion F=ma whereF is force and m mass a acceleration due to gravity i.e. 9.8m/s

Answer 2

(2) 50 kg

Explanation:

Let x be the initial length of the spring when no mass is suspended from it.

we know the relation between the load and deflection in the length of spring as:

$F=k. \Delta x$ ..........................(1)

According to first case:

mass, $m_1=20\ kg$

total length after deflection, $x+\Delta x_1=0.1\ m$

Equation (1) becomes:

$20\times 10=k(0.10-x)$

$200=0.1k-k.x$ .....................................(2)

According to second case:

mass, $m_2=32\ kg$

total length after deflection, $x+\Delta x_2=0.12\ m$

Equation (1) becomes:

$32\times 10=k(0.12-x)$

$320=0.12k-k.x$ .....................................................(3)

Subtract eq. (2) from eq. (3)

$0.02k=120$

$k=6000\ N.m^{-1}$

Now putting this value of k in eq. (2)

$200=6000(0.1-x)$

$x=\frac{1}{15}\ m$

which is the length of unloaded spring.

Now for the length to be 0.15 m:

we need a deflection of, $\Delta x_3=(0.15-\frac{1}{15} )\ m$

putting respective values in eq. (1)

$m_3\times g=k.\Delta x_3$

$m_3\times 10=6000\times (0.15-\frac{1}{15} )$

$m_3=50\ kg$

TOPIC: spring force, deflection of springs

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