Respuesta :
Answer:0.10283 J
Explanation:
Given
mass of block(m)=0.0475 kg
radius of track (r)=0.425 m
when the Block is at Bottom Normal has a magnitude of 3.95 N
Force acting on block at bottom
[tex]N-mg=\frac{mu^2}{r}[/tex]
[tex]N=mg+\frac{mu^2}{r}[/tex]
[tex]3.95=0.0475\timess 9.81+\frac{0.0475\times u^2}{0.425}[/tex]
[tex]u^2=31.172[/tex]
[tex]u=\sqrt{31.172}[/tex]
u=5.583 m/s
At top point
[tex]N+mg=\frac{mv^2}{r}[/tex]
[tex]0.670+0.0475\timess 9.81=\frac{mv^2}{r}[/tex]
[tex]v^2=10.1639[/tex]
v=3.188 m/s
Using Energy conservation
[tex]\frac{mu^2}{2}=\frac{mv^2}{2}+mg\left ( 2r\right )+W_f[/tex]
[tex]W_f=0.4989-0.39607[/tex]
[tex]W_f=0.10283 J[/tex]
i.e. 0.10283 J of energy is wasted while moving up.
The work done on the block by friction during the motion of the block from point A to point B is 0.102 J.
Velocity of the block at top circle
The velocity of the block at the top of the vertical circle is calculated as follows;
[tex]W + mg = \frac{mv^2}{r} \\\\0.67 + 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\1.136 = 0.112 v^2\\\\v^2 = 10.14\\\\v = 3.18 \ m/s[/tex]
Velocity of the block at bottom circle
The velocity of the block at the bottom of the vertical circle is calculated as follows;
[tex]W - mg = \frac{mv^2}{r} \\\\3.95 - 0.0475(9.8)= \frac{0.0475v^2}{0.425} \\\\3.485 = 0.112 v^2\\\\v^2 = 31.12\\\\v = 5.58 \ m/s[/tex]
Work done by friction
The work done by friction is the change in the kinetic energy of the block.
[tex]W _f =P.E_f - \Delta K.E \\\\W_f = mgh - \frac{1}{2} m(v_f^2 - v_i^2)\\\\Wf = 0.0475\times 9.8(2 \times 0.425) - \frac{1}{2} \times 0.0475(5.58^2 - 3.18^2)\\\\W_f = 0.396 - 0.498\\\\W_f = -0.102 \ J[/tex]
Thus, the work done on the block by friction during the motion of the block from point A to point B is 0.102 J.
Learn more about work done by friction here: https://brainly.com/question/12878672
