A person pushes a 16.0-kg shopping cart at a constant velocity for a distance of 22.0 m. She pushes in a direction 29.0° below the horizontal. A 48.0-N frictional force opposes the motion of the cart.
To solve this problem, we can break down the forces acting on the shopping cart into horizontal and vertical components.
The weight of the shopping cart acts downward, and its magnitude can be found using the equation: weight = mass × gravity
weight = 16.0 kg × 9.8 m/s^2 = 156.8 N
The normal force, which is the force exerted by the ground on the shopping cart in the vertical direction, has the same magnitude but opposite direction as the weight of the cart.
The frictional force opposes the motion of the cart and acts in the direction opposite to the applied force. Its magnitude is given as 48.0 N.
Since the cart is moving at a constant velocity, the net force acting on it is zero. Therefore, the sum of the horizontal forces must be zero.
The horizontal forces acting on the cart are the horizontal component of the applied force and the frictional force. Let's assume the applied force is F.
Summing the horizontal forces, we have:
F_cos(29°) - Frictional force = 0
F_cos(29°) - 48.0 N = 0
Rearranging the equation, we find:
F_cos(29°) = 48.0 N
F = 48.0 N / cos(29°) ≈ 57.0 N
Therefore, the magnitude of the applied force is approximately 57.0 N.
Next, we can calculate the work done by the applied force. The work done by a constant force is given by the equation: work = force × distance × cos(angle)
Let's calculate the work done by the applied force:
work = 57.0 N × 22.0 m × cos(29°)
The work done by the applied force is approximately 1100 J.
Since the cart is moving at a constant velocity, the work done by the applied force is equal to the work done by the frictional force. Therefore, the work done by the frictional force is also 1100 J.
The work done by a frictional force is given by the equation: work = force × distance
1100 J = 48.0 N × distance
distance = 1100 J / 48.0 N
The distance traveled by the cart is approximately 22.9 m.