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djyliett [7]
3 years ago
8

An object of mass 9.00 kg attached to an ideal massless spring is pulled with a steady horizontal force across a frictionless le

vel surface. If the spring constant is 95.0 N/m and the spring is stretched by 10.0 cm , what is the magnitude of the acceleration of the object?
Physics
1 answer:
frosja888 [35]3 years ago
5 0

Answer:

Acceleration of the object will be 1.05m/sec^2

Explanation:

We have given mass of the object m = 9 kg

spring constant k = 95 N/m

Spring is stretched by 10 cm

So x = 10 cm = 0.1 m

We know that force is given by F=kx=95\times 0.1=9.5N

From newton's law we also know that force is given by

F = ma , here m is mass and a is acceleration

So 9.5=9\times a

a=1.05m/sec^2

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A spring has a equilibrium length of 10.0 cm. When a force of 40.0 N is applied to the spring, the spring has a length of 14.0 c
mote1985 [20]

Answer:

The value of the spring constant of this spring is 1000 N/m

Explanation:

Given;

equilibrium length of the spring, L = 10.0 cm

new length of the spring, L₀ = 14 cm

applied force on the spring, F = 40 N

extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm

From Hook's law

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