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Ivan
2 years ago
7

A box of mass 1 kg is hung from a spring scale. The reading on the spring scale is approximately 10N. What would be the reading

on the spring scale if a box of mass 5 kg is hung from it? (Assume acceleration due to gravity to be 10 m/s2.)
A) 5 N
B) 10 N
C) 50 N
D) 500 N
Physics
2 answers:
8090 [49]2 years ago
8 0
The answer will be 50N.
This is because the spring reads weight and weight is mass times acceleration due to gravity.5kg*10m/s2=50N
Blababa [14]2 years ago
6 0

Explanation:

It is given that,

When mass is, m₁ = 1 kg

The reading on the spring scale, F = 10 N

If the mass is, m₂ = 5 kg

We need to find the reading on the spring scale. The force of gravity is acting on the box which is given by :  

F=mg

F=5\ kg\times 10\ m/s^2

F = 50 N

So, the reading on the spring scale is 50 N. Hence, this is the required solution.                            

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Arrow_forward
garri49 [273]

Explanation:

(a) Hooke's law:

F = kx

7.50 N = k (0.0300 m)

k = 250 N/m

(b) Angular frequency:

ω = √(k/m)

ω = √((250 N/m) / (0.500 kg))

ω = 22.4 rad/s

Frequency:

f = ω / (2π)

f = 3.56 cycles/s

Period:

T = 1/f

T = 0.281 s

(c) EE = ½ kx²

EE = ½ (250 N/m) (0.0500 m)²

EE = 0.313 J

(d) A = 0.0500 m

(e) vmax = Aω

vmax = (0.0500 m) (22.4 rad/s)

vmax = 1.12 m/s

amax = Aω²

amax = (0.0500 m) (22.4 rad/s)²

amax = 25.0 m/s²

(f) x = A cos(ωt)

x = (0.0500 m) cos(22.4 rad/s × 0.500 s)

x = 0.00919 m

(g) v = dx/dt = -Aω sin(ωt)

v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)

v = -1.10 m/s

a = dv/dt = -Aω² cos(ωt)

a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)

a = -4.59 m/s²

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pantera1 [17]

the answer is in fact True

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