A person pushes a 11.3-kg shopping cart at a constant velocity for a distance of 19.4 m on a flat horizontal surface. She pushes in a direction 29.2 ° below the horizontal. A 44.6-N frictional force opposes the motion of the cart. (a) What is the magnitude of the force that the shopper exerts?

Given: m=11.3kg. A = 29.2 deg., Ff = 44.6N.

Wc = mg = 11.3kg * 9.8N/kg = 110.74N. =
Weight of cart.
Fc = (110.74N,0deg.).
Fp = 110.74sin(0) = 0 = Force parallel to surface.
Fv = 110.74cos(0) = 110.74N. = Force
perpendicular to surface.

Ff = 44.6N = Force of fiction.

Fn=Fap*cos29.2 - Fp - Ff = ma = 0, a=0.
Fap*cos29.2 - 0 - 44.6 = 0,
Fap*cos29.2 = 44.6,
Fap = 44.6 / cos29.2 = 51.1N. = Force
applied.

To find the magnitude of the force that the shopper exerts on the shopping cart, we need to consider the forces acting on the cart and apply Newton's second law of motion.

1. Identify the forces acting on the cart:
- The force exerted by the shopper (unknown magnitude)
- The force of friction opposing motion (44.6 N)

2. Break down the shopper's force into horizontal and vertical components:
- The horizontal component of force is responsible for the cart's constant velocity.
- The vertical component of force does not affect the motion of the cart on a flat surface.

3. Calculate the horizontal and vertical components of the shopper's force:
- The horizontal component is given by F_h = F_shop * cos(θ), where θ is the angle below the horizontal (29.2°).
- The vertical component is given by F_v = F_shop * sin(θ).

4. Since the cart is moving at a constant velocity, the net force acting on the cart in the horizontal direction is zero. Therefore, the horizontal component of the force exerted by the shopper balances the frictional force:
F_h = 44.6 N

5. Substitute the values into the equation:
F_shop * cos(29.2°) = 44.6 N

6. Solve for F_shop, the magnitude of the force that the shopper exerts:
F_shop = 44.6 N / cos(29.2°)

7. Calculate the magnitude of the force exerted by the shopper:
F_shop ≈ 51.8 N

Therefore, the magnitude of the force that the shopper exerts on the shopping cart is approximately 51.8 N.