Answer: the speed is 0.88m/s
Explanation:
The mass is 0.75kg, and the velocity is 2m/s.
Now, for a distance of 1 meter, the block experiences a force of friction, that can be described by:
F = m*g*k, that is in the opposite direction to the movement of the block, and where k is the coefficient of kinetic friction, so we have that the force is:
F = -0.75*9.8*0.17N = -1.2N
Then the acceleration of the block is:
a = F/m = -1.2/0.75 m/s^2 = -1.6m/s^2
Now, the velocity equation will be:
v(t) = (-1.6m/s^2)*t + 2m/s
the position equation will be:
p(t) = (1/2)*(-1.6 m/s^2)*t^2 + 2m/s*t
when p(t) = 1m, the force of friction stops, so we must find this time and then put it in the equation of the velocity:
1 = (-0.8)*t^2 + 2*t
-0.8t^2 + 2t - 1 = 0
the roots can be obtained by:
t = (-2 +- √(2*2 - 4*-1*-0.8))/2*(-0.8) = (-2 +- 0.89)/(-1.6)
then the roots are:
t = (-2 + 0.89)/(-1.6) = 0.7s
t = (-2 - 0.89)/(-1.6) = 1.8s
Here we must choose the smallest positive number, because once the block gets out of the rough area, the acceleration stops.
now, we put that time in the velocity equation:
v(0.7s) = (-1.6m/s^2)*0.7 + 2m/s = 0.88m/s
Answer:
The speed of the block after it passes across the rough surface is [tex]v_{f}=0.82\frac{m}{s}[/tex] .
Explanation:
We are told that the mass of the block is [tex]m=0.75\ kg[/tex], the initial speed is [tex]v_{i}=2\ \frac{m}{s}[/tex], the length of the rough area is [tex]\Delta x=1.0\ m[/tex] and the coefficient of kinetic friction is [tex]\mu_{d}=0.17[/tex].
Because the block doesn't move in the vertical direction the sum of forces is:
[tex]F_{normal} -F_{weight}=0\ \Longrightarrow\ F_{normal} =F_{weight}=m.g[/tex]
[tex]F_{normal} =m.g=0.75\ kg.\ 9.8\frac{m}{s^{2}}=7.35N[/tex]
[tex]F_{normal}=7.35N[/tex]
The force of kinetic friction between the block and the rough surface is:
[tex]F_{friction}=\mu_{d}.F_{normal}=0.17.\ 7.35N=1.25N[/tex]
[tex]F_{friction}=1.25N[/tex]
The work done by the force of kinetic friction is:
[tex]W=-F_{friction}.\Delta x=-1.25N.\ 1.0m=-1.25N.m[/tex]
[tex]W=-1.25J[/tex]
We calculate the speed of the block using that the change in kinetic energy is equal to the work done by the force of kinetic friction:
[tex]W=\Delta K[/tex]
[tex]W=\frac{1}{2}\ m\ v_{f} ^{2} -\frac{1}{2}\ m\ v_{i} ^{2}[/tex]
[tex]-1.25J=\frac{1}{2}\ 0.75kg (v_{f} ^{2}-v_{i} ^{2})[/tex]
[tex]\frac{-2.\ 1.25J}{0.75kg}=v_{f} ^{2}-(2\frac{m}{s})^{2}[/tex]
[tex]-3.33\ \frac{m^{2}}{s^{2}}=v_{f}^{2}-4\ \frac{m^{2}}{s^{2}}[/tex]
[tex]v_{f} ^{2}=-3.33\ \frac{m^{2}}{s^{2}}+4\ \frac{m^{2}}{s^{2}}=0.67\ \frac{m^{2}}{s^{2}}[/tex]
[tex]v_{f}=0.82\ \frac{m}{s}[/tex]
