Answer
given,
mass of the block = 200 g = 0.2 Kg
Velocity at A = 0 m/s
Velocity at B = 8 m/s
slide to the horizontal distance = 10 m
height of the block be = 4 m
potential energy of the block
P = m g h
P = 0.2 x 9.8 x 4
P =7.84 J
kinetic energy
[tex]KE = \dfrac{1}{2}mv^2[/tex]
[tex]KE = \dfrac{1}{2}\times 0.2 \times 7.84^2[/tex]
[tex]KE =6.14 J[/tex]
Work = P - KE
work = 7.84 - 6.14
work = 1.7 J
b) v² = u² + 2 a s
0 = 8² - 2 x a x 10
a = 3.2 m/s²
ma - μ mg = 0
[tex]\mu = \dfrac{a}{g}[/tex]
[tex]\mu = \dfrac{3.2}{9.8}[/tex]
[tex]\mu = 0.327[/tex]
The work done by friction along the horizontal surface is 6.4 J, and the coefficient of kinetic friction along the horizontal surface is 0.327.
The normal force on the block is calculated as;
[tex]F_n = mg[/tex]
[tex]F_n = 0.2 \times 9.8\\\\F_n = 1.96 \ N[/tex]
The acceleration of the block over the given distance is calculated as;
[tex]v^2 = u^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\a = \frac{v^2}{2s} \\\\a = \frac{(8)^2}{2\times 10} \\\\a = 3.2 \ m/s^2[/tex]
The coefficient of kinetic friction is calculated as;
a = μg
[tex]\mu_k = \frac{a}{g} \\\\\mu_k =\frac{3.2}{9.8} \\\\\mu_k = 0.327[/tex]
The frictional force along the surface is calculated as;
[tex]F_k = \mu_k F_n\\\\F_k = 0.327\times 1.96\\\\F_k = 0.64 \ N[/tex]
The work done by friction along the horizontal surface is calculated as;
[tex]Work-done = F_k \times d\\\\Work-done = 0.64 \times 10\\\\Work-done = 6.4 \ J[/tex]
Learn more here:https://brainly.com/question/14268613
