Respuesta :
Answer:
6m
Explanation:
The initial potential energy of the block P.E =mgh
where mass m = 3kg
g= 9.81 m/s2
h = 2m
Hence P.E = 3*9.81*2 = 58.8 Joules
This energy converts to kinetic energy when the block is released from the top
K.E = [tex] mv^2/2 = 58.8 [/tex]
Solving for v, we obtain 6.26 m/s
The acceleration of the block by Newton's second law of motion,
ma = (μk)*mg
where μk = 0.1
Hence, a = 0.1*9.81 = 0.98m/s/s
Applying Newton's second equation of motion to solve,
[tex] v^2 - u^2 =2as [/tex]
where u = 0 m/s (released from rest0
Hence, [tex] 6.26^2 = 2*0.98*s [/tex]
Solving for s,
s= 20m
Therefore, the distance after two complete cycles = 20 -2(7) = 6m
The expressions that describes where the block stops from the left end of the horizontal surface, marked x = 0 is; [tex]x = \dfrac{0.3 \cdot R}{\mu_k}[/tex]
The reasons why the above values are correct are as follows;
The given parameters are;
Mass of the block, m = 3 kg
Radius of the quarter circular track, R = 2 m
Length of the horizontal track, L = 7 m
Coefficient of friction between the block and the horizontal surface, [tex]\mu_{k}[/tex] = 0.1
Required:
To form an expression that describes where the block stops as measured from the left end of the horizontal surface marked x = 0
Solution;
A quarter of a circle has the radii lengths arranged at right angles to each other, both of which are perpendicular to tangents of the circle
A flat surface, such as the horizontal surface in the question is a tangent to the circle
Therefore;
The radius of the circle, perpendicular to the horizontal surface = The height from which the block is released
Potential energy = m·g·h
Where;
g = Acceleration due to gravity ≈ 9.81 m/s²
Therefore, the potential energy of the block, P.E., is given as follows;
P.E. = m·g·R
P.E. = 3 kg × 2 m × 9.81 m/s² = 58.86 J
The potential energy of the block, P.E. = 58.86 J
The frictional force due to the block, [tex]F_f[/tex] = m·g·[tex]\mu_{k}[/tex]
3 kg × 9.81 m/s²× 0.1 = 2.943 N
Work done by the friction as the block moves along the horizontal surface, is given as follows;
Work done = Force × Distance moved in the direction of the force
∴ Work done by friction = [tex]F_f[/tex] × d
Where;
d = The distance from the left end where the block reaches
Work done by friction also = Energy transferred = ΔE
When the block stops, the total energy in the block is transferred
Therefore, when the block stops, according to the law of conservation of energy;
Work done by friction = [tex]F_f[/tex] × d = Energy transferred = ΔE = P.E.
2.943·d = 58.86
[tex]d = \dfrac{m \cdot g \cdot R}{m \cdot g \cdot \mu_k} = \dfrac{R}{\mu_k}[/tex]
[tex]d = \dfrac{58.86 \, J}{2.943 \, N } = 20 \, m[/tex]
Therefore, given that work is done by friction only on the horizontal plane, we have;
The distance travelled on the horizontal surface = 20 m
Therefore, the block traverses the horizontal surface twice, and then for another 6 meters
Let N represent the number of times the block traverses on the horizontal surface, and let L represent the length of the horizontal surface, we have;
d = N·L + x (when N is even)
d = N·L + L - x = L·(N + 1) - x (when N is odd)
[tex]\dfrac{R}{\mu_k} = N \cdot L + x[/tex]
[tex]\dfrac{R}{\mu_k} - N \cdot L = x[/tex]
[tex]x = L \cdot (N + 1) - \dfrac{R}{\mu_k}[/tex]
Therefore;
The expressions that describe where the block stops as measured from the left end of the horizontal surface marked x = 0 are;
[tex]\dfrac{R}{\mu_k} - 7 \cdot N_{even} = x[/tex] and [tex]x = 7 \cdot(N_{odd} + 1) - \dfrac{R}{\mu_k}[/tex]
The distance where the block stops from the left end is 6 meters, therefore, we have;
[tex]\dfrac{R}{\mu_k} = 20[/tex]
[tex]x = 6 = \dfrac{0.3 \cdot R}{\mu_k}[/tex]
Learn more about the law of conservation of energy here:
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