(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.

Question

(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.

Answer 1

To find the force of friction acting on the filing cabinet, we can use the formula:Force of friction = coefficient of friction * normal forceThe normal force is equal to the weight of the cabinet, which is the product of its mass and the acceleration due to gravity (g = 9.8 m/s²).Normal force = mass * acceleration due to gravityNormal force = 85 kg * 9.8 m/s²Normal force = 833.8 NNow, we can calculate the force of friction:Force of friction = coefficient of friction * normal forceForce of friction = 0.22 * 833.8 NForce of friction = 183.836 NTherefore, the force of friction acting on the filing cabinet is approximately 183.8 N.

Вопрос

(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.

Решения

Ответ