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

To identify the forces acting on the cabinet and calculate their magnitudes, we need to consider the following forces:<br /><br />1. Gravitational force (weight) acting downward on the cabinet.<br />2. Normal force acting upward on the cabinet.<br />3. Frictional force acting horizontally opposing the motion of the cabinet.<br /><br />Given information:<br />- Mass of the cabinet, m = 85 kg<br />- Distance, d = 3.1 m<br />- Coefficient of kinetic friction, μk = 0.22<br /><br />(i) Forces acting on the cabinet:<br /><br />1. Gravitational force (weight):<br /> 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).<br /> Gravitational force = m × g = 85 kg × 9.8 m/s² = 833.8 N<br /><br />2. Normal force:<br /> The normal force is equal to the gravitational force acting on the cabinet, as there is no vertical motion.<br /> Normal force = 833.8 N<br /><br />3. Frictional force:<br /> The frictional force is equal to the product of the coefficient of kinetic friction and the normal force.<br /> Frictional force = μk × Normal force = 0.22 × 833.8 N = 183.836 N<br /><br />Therefore, the forces acting on the cabinet are:<br />- Gravitational force (weight) = 833.8 N<br />- Normal force = 833.8 N<br />- Frictional force = 183.836 N