Question

An object attached to an ideal massless spring is pulled across a frictionless surface. If the spring constant is 45 N/m and the spring is stretched by 0.88 m when the object is accelerating at 0.8 m/s2, what is the mass of the object

Answer 1

Answer:

The mass of the object is 49.5kg which is approximately 50kg

Explanation:

Given that

Spring constant (k)=45N/m

The extension (e)=0.88m

Also given that the acceleration is 0.8m/s²

Force by the spring is given as

Using hooke's law

According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,

F = ke where

F is the applied force

k is the spring constant

e is the extension

From the formula k = F/e

F=ke

m is the mass of the block = ?

a is the acceleration = 0.8m/s²

e is the extension of the spring =  0.88m

k is the spring constant = 45N/m

F=45×0.88

F=39.6N

Now this force will set the object in motion, now using newton second law of motion

F=ma

Then, m=F/a

m=39.6/0.8

m=49.5kg

The mass of the object is 49.5kg which is approximately 50kg

Answer 2

Answer:

Explanation:

Using Hooke's law,

F = -k × delta x

Where,

F = force

k = spring constant

= 45 N/m

Delta x = spring displacement

= 0.88 m

F = 45 × 0.88

= 39.6 N

a = 0.8 m/s^2

Force, F = Mass, M × acceleration, a

M = 39.6/0.8

= 49.5 kg