The initial kinetic energy of the block is 0 J.
The final kinetic energy of the block is 4.9 J.
The change in potential energy is -5.88 J.
The work done by normal force is 0 J.
The work done by frictional force is 0.98 J.
The magnitude of frictional force is 0.196 N.
The coefficient of kinetic friction is 0.1.
The normal force is greater than frictional force.
Initial kinetic energy of the block
The initial kinetic energy of the block is calculated as follows;
K.Ei = ¹/₂mvi²
K.Ei = 0.5(0.2)(0²)
K.Ei = 0 J
Final kinetic energy of the block
K.Ef = ¹/₂mvf²
where;
K.Ef = ¹/₂(0.2)(7²)
K.Ef = 4.9 J
Change in potential energy
ΔP.E = P.Ef - P.Ei
ΔP.E = mghf - mghi
ΔP.E = mg(hf - hi)
ΔP.E = 0.2 x 9.8(0 - 3)
ΔP.E = -5.88 J
Work done by normal force
Wn = fd cosθ
Wn = mgl cosθ
where;
- θ angle of inclination of the normal force = 90⁰
Wn = 0.2 x 9.8 x 5 x cos(90)
Wn = 0 J
Work done by frictional force
Wf = P.E - K.E
Wf = 5.88 J - 4.9 J
Wf = 0.98 J
Magnitude of frictional force
Wf = fd
where;
- f is the frictional force
- d is the distance moved by the block
f = Wf/d
f = 0.98 J / 5 m
f = 0.196 N
Coefficient of kinetic friction
f = μmg
μ = f/mg
μ = 0.196/(0.2 x 9.8)
μ = 0.1
Angle of inclination of the plane
sin θ = H/L
sin θ = 3/5
sin θ = 0.6
θ = arc sin (0.6)
θ = 37⁰
Normal force
Fn = mg cosθ
Fn = 0.2 x 9.8 x cos(37)
Fn = 1.56 N
Thus, the normal force is greater than frictional force.
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