Question

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A block of mass `M` slides along a horizontal table with speed `v_0`. At `x=0`, it hits a spring with spring constant `k` and begins to experience a frictional force. The coefficient of friction is given by `mu=bx` {b is a positive constant}. What is the loss in mechanical energy when the block first comes momentarily to rest?

Answer 1

Initial kinetic energy of the block is

`KE =1/2Mv_0^2`

Using the Law of Conservation of energy, the block comes to rest when the mechanical energy stored in the compressed spring and work done against the frictional force is equal to this value.
Let the block first come momentarily to rest at `x=d`

Mechanical energy stored in the compressed spring `= 1/2 k d^2`

We know that force of friction `F_f` is given as

`F_f = muN=(bx)Mg`
where `N` is the normal reaction and `g` is gravity.

Energy lost is Work done against friction when block moves from `x=0` to `x=d`. This is found with the help of integral of `vecFcdotvec(dx)`.

`W_f=int_0^d\ bMg\ x\ dx`
`W_f=1/2bMgd^2`

Now we have `1/2Mv_0^2=1/2 k d^2+1/2bMgd^2`
Solving for `d^2` we get

`Mv_0^2= (k +bMg)d^2`
`=> d^2=(Mv_0^2)/(k +bMg)`

Inserting in (2) we get
Energy lost`=1/2bMgxx(Mv_0^2)/(k +bMg)=1/2(bM^2g)/(k +bMg)v_0^2`

Answer 2

`1/2(b m^2 g)/(k + b m g) v_0^2`

Explanation:

`1/2mv_0^2= 1/2k x_0^2+ int_0^(x_0) bx m g dx` or

`1/2mv_0^2= 1/2k x_0^2+ 1/2 bx_0^2 m g` solving for `x_0^2`

`x_0^2= m/(k+bmg) v_0^2`

now substituting into

`1/2 bx_0^2 m g`

we get

losses due to friction `= 1/2(b m^2 g)/(k + b m g) v_0^2`