A car with a mass of 2 tons starts off with an acceleration of 2 m / s squared and accelerates for 5 s

A car with a mass of 2 tons starts off with an acceleration of 2 m / s squared and accelerates for 5 s on a horizontal path. What work is done during this time if the coefficient of resistance to movement is 0.01

To determine the amount of work performed by the presented car in 5 seconds of acceleration, we apply the formula: A = F * S = (m * a + Ftr) * 0.5 * a * t ^ 2 = (m * a + μ * m * g ) * 0.5 * a * t ^ 2 = (a + μ * g) * 0.5 * a * t ^ 2 * m.

Data: a – acceleration during acceleration (a = 2 m / s2); μ – coeff. resistance to movement (μ = 0.01); g – acceleration due to gravity (g ≈ 10 m / s2); t – acceleration duration (t = 5 s); m is the mass of the presented car (m = 2 t = 2 * 10 ^ 3 kg).

Calculation: A = (a + μ * g) * 0.5 * a * t ^ 2 * m = (2 + 0.01 * 10) * 0.5 * 2 * 5 ^ 2 * 2 * 10 ^ 3 = 105 * 10 ^ 3 J.

Answer: For 5 seconds of acceleration, the presented car performed work of 105 kJ.