Question

Определить работу, совершенную двигателем автомобиля при разгоне, если масса автомобиля 1,5 т и он движется с ускорением 2 м/с^2 в течение 10 с. Коэффицент трения 0, 02 Пожалуйста, сделайте всё с решением и с ответом. Заранее благодарю! ;)))

Answer 1

сила трения  Fc=mgk

равнодействующая R=ma

равнодействующая  R=F-Fc

сила тяги двигателя F=R+Fc = ma + mgk = m (a+gk)

g=10 м/с2

t =10c

Vo=0

пройденное расстояние  S=Vot +at^2/2 = at^2/2

работа  A=F*S =m (a+gk) *at^2/2 =1500(2+10*0.02)*2*10^2/2 =330 000 Дж = 330 кДж

 

ОТВЕТ 330 кДж

Answer 2

Calculation of Work Done by the Car's Engine

To determine the work done by the car's engine during acceleration, we need to calculate the force exerted by the engine and the distance over which this force is applied.

Given information: - Mass of the car (m) = 1.5 t = 1500 kg - Acceleration (a) = 2 m/s^2 - Time (t) = 10 s - Coefficient of friction (μ) = 0.02

To calculate the force exerted by the engine, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a):

F = m * a

Substituting the given values: F = 1500 kg * 2 m/s^2 = 3000 N

Now, we need to calculate the distance over which this force is applied. Since the car is accelerating, we can use the equation of motion:

s = ut + (1/2)at^2

Where: - s is the distance - u is the initial velocity (assumed to be 0 since the car is starting from rest) - a is the acceleration - t is the time

Substituting the given values: s = 0 + (1/2) * 2 m/s^2 * (10 s)^2 = 100 m

Now, we can calculate the work done by the engine using the formula:

Work (W) = Force (F) * Distance (s)

Substituting the calculated values: W = 3000 N * 100 m = 300,000 J

Therefore, the work done by the car's engine during acceleration is 300,000 Joules (J).

Please note that the above calculation assumes that the only force acting on the car is the force exerted by the engine. In reality, there may be other forces involved, such as air resistance and friction. However, since the coefficient of friction was given as 0.02, we can assume that the effect of friction is already considered in the given values.