Question

g 16.0 kg sled is pulled on a horizontal frictionless ice rink with a light spring of force constant 220 N/m. If the sled has an acceleration of 2.0 when pulled horizontally, how much does the spring stretch (in meters)

Answer 1

`x=-0.145m`

Step-by-step explanation:

Information we have:

Mass of the sled:
`m=16kg`

Spring constant:
`k=220N/m`

acceleration of the sled:
`a=2m/s^2`

We use the formula for the force on a spring (Hooke's Law)

`F=-kx`

where k is the spring constant and x the distance the mass moved from the equilibrium

and we also use Newton's second Law of motion:

`F=ma`

we combine these two equations:

`-kx=ma`

we solve the last equation for the distance x:

`x=-(ma)/(k)`

and substitute the values:

`x=-((16kg)(2m/s^2))/(220N/m)\\ x=-0.145m`

the negative number means that the mass was moved in the opposite direction that the force, this because the force in a spring is restorative and points towards the equilibrium point

Answer 2

Spring stretches by 0.145m

Step-by-step explanation:

We are given;

Mass of sled;m = 16 kg

Force constant;k = 220 N/m

Acceleration;a = 2 m/s²

Now, we know that from Newton's second law of motion, Force is given by the expression ; F = ma

Thus,

F = 16 x 2

F = 32 N

Now from Hooke's law, we know that this force is expressed as;

F = -kx

Where x is the distance that the spring stretches.

The minus sign shows that this force is in the opposite direction of the force that’s stretching or compressing the spring. But we'll take the magnitude and therefore ignore the negative sign to get F = Kx

Thus,

32 = 220x

x = 32/220

x = 0.145 m