A body weighing 5 kg slid off an inclined plane 2 m long. Find the work of the force of gravity, the reaction force of the support and the friction force. The angle of inclination of the plane is 30 degrees, the coefficient of friction is 0.35
m = 5 kg.
g = 10 m / s2.
S = 2 m.
μ = 0.35.
∠α = 30 °.
Аmg = m * g * cos (90 ° – ∠α).
Аmg = 5 kg * 10 m / s2 * cos (90 ° – 30 °) = 25 J.
AFtr = Ftr * S = μ * m * g * cosα * S.
AFtr = 0.35 * 5 kg * 10 m / s2 * cos30 ° * 2 m = 30 J.
АN = N * S * cos0 ° = 0.
Answer: Amg = 25 J, AFtr = 30 J, AN = 0.
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