Вопрос
Q2 (a) A clerk pushes a filing cabinet, whose mass m is 85 kg, at a constant speed across a tiled floor for a distance d of 3 .1 m. the coefficient of friction u_(k) between the bottom of the cabinet and the carpet is 0.22. (i) Identify the forces that act on the cabinet and calculate their
Решения
4.4
(198 Голоса)
Эдгар
ветеран · Репетитор 10 лет
Ответ
To identify the forces acting on the cabinet and calculate their magnitudes, we need to consider the following forces:1. Gravitational force (weight) acting downward on the cabinet.2. Normal force acting upward on the cabinet.3. Frictional force acting horizontally opposing the motion of the cabinet.Given information:- Mass of the cabinet, m = 85 kg- Distance, d = 3.1 m- Coefficient of kinetic friction, μk = 0.22(i) Forces acting on the cabinet:1. Gravitational force (weight): The gravitational force acting on the cabinet is equal to the product of its mass and the acceleration due to gravity (g = 9.8 m/s²). Gravitational force = m × g = 85 kg × 9.8 m/s² = 833.8 N2. Normal force: The normal force is equal to the gravitational force acting on the cabinet, as there is no vertical motion. Normal force = 833.8 N3. Frictional force: The frictional force is equal to the product of the coefficient of kinetic friction and the normal force. Frictional force = μk × Normal force = 0.22 × 833.8 N = 183.836 NTherefore, the forces acting on the cabinet are:- Gravitational force (weight) = 833.8 N- Normal force = 833.8 N- Frictional force = 183.836 N