Question

Бак массой m1=2 кг покоится на тележке массой m2=10 кг, которую разгоняют с ускорением a=5 м/с2. Коэффициент трения между баком и тележкой 0.6. Определите силу трения, действующую на бак со стороны тележки.

Answer 1

Максимальная сила трения меж баком и тележкой Fтр=μN=μmg=0,6•2•10=12 H. Со стороны тележки действует сила трения на бак, сообщающая ему ускорение F=ma=2•5=10 Н, что меньше 12 Н. Значит бак покоился на тележке, двигался вместе с ней.

Answer 2

#### Problem Analysis We are given the following information: - Mass of the tank, m1 = 2 kg - Mass of the cart, m2 = 10 kg - Acceleration of the cart, a = 5 m/s^2 - Coefficient of friction between the tank and the cart, μ = 0.6 We need to determine the frictional force acting on the tank from the cart. #### Solution To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. We can also consider the frictional force between the tank and the cart. The net force acting on the tank is given by: **F_net = m1 * a** The frictional force between the tank and the cart is given by: **F_friction = μ * (m1 + m2) * g** where g is the acceleration due to gravity. #### Calculation Let's calculate the net force and the frictional force. Given: - m1 = 2 kg - m2 = 10 kg - a = 5 m/s^2 - μ = 0.6 - g = 9.8 m/s^2 (acceleration due to gravity) Using the formula for the net force: **F_net = m1 * a** Substituting the given values: **F_net = 2 kg * 5 m/s^2 = 10 N** Using the formula for the frictional force: **F_friction = μ * (m1 + m2) * g** Substituting the given values: **F_friction = 0.6 * (2 kg + 10 kg) * 9.8 m/s^2 = 70.56 N** #### Answer The force of friction acting on the tank from the cart is approximately **70.56 N**.