A person pushes a 25 kg shopping cart with a force of 280 N at an angle of 15 degrees and it has a coefficient of friction of 1.2.

Question

A person pushes a 25 kg shopping cart with a force of 280 N at an angle of 15 degrees and it has a coefficient of friction of 1.2. Find the cart's acceleration, the friction force, the applied force in both the horizontal and vertical directions, the object's weight, and the normal force.

Answer 1

A person pushes a 25 kg shopping cart with a force of 280 N at an angle of 15 degrees and it has a coefficient of friction of 1.2. Find the cart's acceleration, the friction force, the applied force in both the horizontal and vertical directions, the object's weight, and the normal force.

Acceleration of cart = (280Cos 15 - 1.2*280 Sin 15) / 25 As acceleration = Net force/mass

Friction force = 1.2 *280 Sin 15 ( Friction force =Coefficient of friction times Normal reaction)

Horizontal force = 280 Cos 15

Normal force = 280 Sin 15