A block of mass 2.0 kg is given an initial speed along the floor towards a spring as shown. The coefficient of kinetic friction between the floor and the block is 0.4 and force constant of the spring is `5.6xx10^(3)` N/m. The block compresses the spring by 10cm before it stops for a moment. What is the initial speed (m/s) of the block?

Explanation

Because the block is moving, a kinetic frictional force acts on it. The magnitude is given by Eq. 6-2 `(f_(k)=mu_(k)F_(N)`, where `F_(N)` is the normal force). The direction is opposite the motion (the friction opposes the sliding).
Calculating `F_(N)`: Because we need the magnitude `f_(k)` of the frictional force, we first must calculate the magnitude `F_(N)` of the normal force. Figure 6-5b is a free-body diagram showing the forces along the vertical y axis. The normal force is upward, the gravitational force with magnitude mg is downward, and (note) the vertical component `F_(y)` of the applied force is upward. That component is shown in Fig. 6-5c, where we can see that `F_(y)=Fsintheta`. We can write Newton.s second law `(F_("net"=mveca))` for those forces along the y axis as
`F_(N)+Fsintheta-mg=m(0),`

Figure 65 (a) A force is applied to a moving block. (b) The vertical forces. (c) The components of the applied force. (d) The horizontal forces and acceleration.
where we substituted zero for the acceleration along the y axis (the block does not even move along that axis). Thus,
`F_(N)=mg-Fsintheta`.
Calculating acceleration a: Figure 6-5d is a free-body diagram for motion along the x axis. The horizontal component `F_(x)` of the applied force is rightward, from Fig. 6-5c, we see that `F_(x)=Fcostheta`. The frictional force has magnitude `f_(x)(=kF_(N))` and is leftward. Writing Newton.s second law for motion along the x axis gives us
`Fcostheta-mu_(k)F_(N)=ma`
Substituting for `F_(N)` from Eq. 6-15 and solving for a lead to
`a=(F)/(m)costheta-mu_(k)(g-(F)/(m)sintheta)`
Finding a maximum: To find the value of `theta` that maximizes a, we take the derivative of a with respect to `theta` and set the result equal to zero:
`(da)/(d""theta)=-(F)/(m)sintheta+mu_(k)(F)/(m)costheta=0`.
Rearranging and using the identity `(sintheta)//(costheta)=tantheta` give us
`tantheta=mu_(k)`.
Solving for `theta` and substituting the given `mu_(k)=0.40`, we find that the acceleration will be maximum if
`theta=tan^(-1)mu_(k)`
Comment: As we increase `theta` from 0, more of the applied force `vecF` is upward, relieving the normal force. The decrease in the normal force causes a decrease in the frictional force, which opposes the block.s motion. Thus, the block.s acceleration tends to increase. However, the increase in `theta` also decreases the horizontal component of `vecF`, and so the block.s acceleration tends to decrease. These opposing tendencies produce a maximum acceleration at `theta= 22^(@)`