A shopper pushes a 5.32 kg grocery cart with a 12.7 N force directed at -28.7° below horizontal. A friction force of 8.33 N pushes back against the motion. What is the acceleration of the cart?

Answer 1

Since mass of cart is 5.32, its weight is 5.32 x 9.81N

The component of weight acting down the slope is 5.32 x 9.81 x cos(-28.7°) = 45.78N

In addition, the cart is pushed down the slope with a force of 12.7N and is impeded by a frictional force of 8.33N

∴ net force on the cart is 45.78 + 12.7-8.33 = 50.15N

a = F/m ∴ a = 50.15/5.32 = 9.43 m/s²

Mr. A