Question

Help?please

This question is physics

Answer 1

0.191131498.....

Explanation:

For the purpose of simplicity let the acceleration due to the gravity near the earth's surface be `g = 9.81m/s^2`

To find the initial speed with which the object is launched at the INSTANT it leaves the spring,
`k = 200 N/m`
`1/2 kx^2 = 1/2 mv_0^2` where `k` is the spring constant of the spring

cancelling the `1/2`on either side and rewriting to solve for `v`, we get `v_0 = sqrt((kx^2) / m)`
`v_0 = sqrt((200*(0.15)^2)/2)`
`v_0 = sqrt(2.25) = 1.5 m/s`

The general formula for Work is W = Force * distance = `F * d`
To find work done by friction, since it opposes the direction of motion of the block,
`W_f = -mu_k*F*d = -mu_k*m*g*d`

Since the block comes to rest and all the kinetic energy is converted to heat due to friction and other forms of energy,
`-mu_k*m*g*d = -1/2mv_0^2`

Rewriting to solve for `mu_k` gives us
`mu_k = v_0^2/(2*g*d)`
`mu_k = (1.5)^2/(2*9.81*0.6) = 2.25/11.772 ~~ 0.191131498... ~~0.191`

Therefore `mu_k ~~ 0.191`

Answer 2

`0.19`, two significant figures

Explanation:

As the spring is compressed it has mechanical potential energy which is given by the expression

`PE=1/2 kx^2`
where `k` is spring constant and `x` is deformation of the spring.

On releasing of spring, all this energy is transferred to the block as its kinetic energy.

As the block moves on the horizontal surface, its kinetic energy is used to do work against force of friction `vecF`

Now `vecF=mu_kvecN`
where `mu_k` is coefficient of friction and `vec N` is normal reaction`=mvecg`.

Inserting various values we have
`F=mu_kxx2.0xx9.8=19.6mu_k`

We know that work done is `W = vecF * vecs`

In this case `theta` angle between the force of friction and displacement is `0^@=>cos theta=cos0^@=1.`

`:.W = F * s`
`=>W = 19.6mu_k xx 0.60`

Since the block comes to rest this is equal to initial potential energy of the spring. Equating the two we get

`19.6mu_k xx 0.60=1/2kx^2`

Inserting given values we get
`19.6mu_k xx 0.60=1/2xx200xx (0.15)^2`
`=>mu_k =100xx (0.15)^2xx1/(19.6xx0.60)`
`=>mu_k = 0.19`, two significant figures