Question

Троллейбус массой 10 т, трогаясь с места в горку под углом 30 градусов, приобрел на пути 50 м скорость 10 м/с. Найдите коэффициент трения, если сила тяги равна 14 кН. РИСУНОК 100Б

Answer 1

Ответ:

Дано:

m=10⁴

S=50

Δv=10

F=14·10³

Найти:

µ  

Решение:

 

F-Fтр=ma (по II з-ну Н)

Fтр - скольжения ⇒ Fтр=µN, где N - сила реакции опоры, численно равна P(весу), так как a по y равно нулю ⇒ P=N=mg

F-µmg=ma

µ=(F-ma)/mg

 

Найдём a

S=ΔV²/ 2a, отсюда

a=ΔV²/2S, так как v₀=0 ⇒ a=1

 

Найдём µ (при g≈10)

µ=(14·10³-10⁴)/10⁵ = 0,04

 

Ответ: 0,04

Объяснение:

Answer 2

Problem Analysis

We are given the following information: - Mass of the trolleybus: 10 tons - Angle of inclination: 30 degrees - Distance covered: 50 meters - Initial speed: 0 m/s - Final speed: 10 m/s - Traction force: 14 kN

We need to find the coefficient of friction.

Solution

To solve this problem, we can use the concept of work and energy. The work done by the traction force is equal to the change in kinetic energy of the trolleybus.

The work done by the traction force is given by the equation:

Work = Force * Distance * cos(angle)

The change in kinetic energy is given by the equation:

Change in Kinetic Energy = (1/2) * Mass * (Final Velocity^2 - Initial Velocity^2)

Since the initial velocity is 0 m/s, the equation simplifies to:

Change in Kinetic Energy = (1/2) * Mass * Final Velocity^2

Setting the work done by the traction force equal to the change in kinetic energy, we can solve for the coefficient of friction:

Force * Distance * cos(angle) = (1/2) * Mass * Final Velocity^2

Rearranging the equation, we get:

Coefficient of Friction = (Force * Distance * cos(angle)) / (0.5 * Mass * Final Velocity^2)

Now, let's substitute the given values into the equation and calculate the coefficient of friction.

Calculation

Given: - Mass of the trolleybus: 10 tons (1 ton = 1000 kg) - Angle of inclination: 30 degrees - Distance covered: 50 meters - Final speed: 10 m/s - Traction force: 14 kN (1 kN = 1000 N)

Converting the mass of the trolleybus to kilograms: Mass = 10 tons * 1000 kg/ton = 10000 kg

Converting the traction force to Newtons: Force = 14 kN * 1000 N/kN = 14000 N

Substituting the values into the equation: Coefficient of Friction = (14000 N * 50 m * cos(30 degrees)) / (0.5 * 10000 kg * (10 m/s)^2)

Calculating the coefficient of friction using a calculator: Coefficient of Friction ≈ 0.242

Therefore, the coefficient of friction is approximately 0.242.

Answer

The coefficient of friction for the trolleybus is approximately 0.242.