1. on an inclined surface. The surface makes 30⁰ angle with horizon.
An object with 5kg mass is sliding downward on an inclined surface. The surface makes 30⁰ angle with horizon. If the coefficient of kinetic friction is 0.25, what is the object acceleration?
Break it down:
m = 5kg  W = mg = 5 * 9.8 = 49N; the weight of the object
An object with 5kg mass
The object is moving downward, then friction will act in opposite side, upward along the surface.
is sliding downward
on an inclined surface. The surface makes
30⁰ angle with horizon.
θ = 30⁰, the surface deviates 30⁰above horizon.
If the coefficient of kinetic friction is 0.25,
μK = 0.25; kinetic friction coefficient.
what is the object acceleration?
a =? Both magnitude and direction?
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2. An object with 5kg mass is sliding downward on an inclined surface
An object with 5kg mass is sliding downward on an inclined surface. The surface makes 30⁰ angle with horizon. If the coefficient of kinetic friction is 0.25, what is the object acceleration?
Solution:
Draw a diagram
+ y
Assume +x axis along the surface and Y axis perpendicular to the surface.
Moving surface Fk FN
mg sin30⁰
mg cos30⁰ θ
+ x
Horizon
W = mg
θ = 30⁰
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3. FN, always perpendicular to the contact surface
An object with 5kg mass is sliding downward on an inclined surface. The surface makes 30⁰ angle with horizon. If the coefficient of kinetic friction is 0.25, what is the object acceleration?
Solution:
Remember that:
FN, always perpendicular to the contact surface
W (weight force), always perpendicular to the horizon surface
Fk = μKFN, always in opposite direction with the motion direction
Component of W, along the moving surface (x axis) is W sinθ or “mg sin30⁰”
Component of W, perpendicular to the moving surface is W cosθ or “mg cos30⁰”
Weight vector is perpendicular to the horizon. So, the Y axis is perpendicular to the moving surface. The horizon makes an angle θ = 30⁰ with the moving surface. Therefore the weight vector and y axis make an angle of θ = 30⁰ (see the diagram).
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4. Calculate net forces along x and y axes
An object with 5kg mass is sliding downward on an inclined surface. The surface makes 30⁰ angle with horizon. If the coefficient of kinetic friction is 0.25, what is the object acceleration?
Solution:
Calculate net forces along x and y axes
+ mg sin30⁰ + (- FK)
∑Fx =
∑Fy =
The object moves along +X axis
∑Fy = +FN + (-mg cos30⁰)
Write the Newton’s second law of the motion
 + mg sin30⁰ + (- FK) = max
∑Fx = max
∑Fy = may
the object does not move along y axis then ay = 0
 ∑Fy = m (0) = 0
 +FN + (-mg cos30⁰) = 0
 FN = mg cos30⁰
 FN = 5 * 9.8 * 0.87
 FN = 42.4N
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5. Downward along moving surface or + X axis
An object with 5kg mass is sliding downward on an inclined surface. The surface makes 30⁰ angle with horizon. If the coefficient of kinetic friction is 0.25, what is the object acceleration?
Solution:
Calculate Fk
FK = μK FN
 FK = 0.25 * 42.4
 FK = 10.6N
Substitute calculated FK in the equation from Newton’s second law along X axis
∑Fx = max
 + mg sin30⁰ + (- FK) = max
 + [5 * 9.8 * sin30⁰] + (- 10.6) = 5 * ax
 ax = 2.78 m/s2 ≈ 2.8 m/s2
Downward along moving surface or + X axis
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