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### Given Data:
- Mass of the grocery cart \( m = 25 \, \text{kg} \)
- Force applied \( F = 2250 \, \text{N} \)
- Angle of force \( \theta = 60^\circ \) with the horizontal
- Distance \( d = 7.5 \, \text{m} \)
- Coefficient of kinetic friction \( \mu_k = 0.32 \)

### Part (a) – Work Done by the 2250 N Force
The work done by a force \( F \) applied at an angle \( \theta \) over a distance \( d \) is given by:
\[
W = F \cdot d \cdot \cos(\theta)
\]
Substituting the given values:
\[
W = 2250 \cdot 7.5 \cdot \cos(60^\circ)
\]
Since \( \cos(60^\circ) = 0.5 \), the work becomes:
\[
W = 2250 \cdot 7.5 \cdot 0.5 = 8437.5 \, \text{J}
\]
So, the work done by the force is \( 8437.5 \, \text{J} \).

### Part (b) – Work Done by Friction
The frictional force is given by:
\[
f_k = \mu_k \cdot N
\]
where \( N \) is the normal force. The normal force can be found by considering the forces in the vertical direction.

In the vertical direction, we have:
\[
N + F \cdot \sin(\theta) = mg
\]
Solving for \( N \):
\[
N = mg - F \cdot \sin(\theta)
\]
Substituting the known values:
\[
N = (25 \cdot 9.81) - 2250 \cdot \sin(60^\circ)
\]
Using \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \):
\[
N = 245.25 - 2250 \cdot 0.866 = 245.25 - 1948.5 = -1703.25 \, \text{N}
\]
The negative value suggests that the vertical component of the applied force is so large that it exceeds the weight of the cart, making the frictional force irrelevant. This could mean the cart is lifted off the ground, depending on interpretation.

Would you like further clarification or more details?

Here are 5 related questions to explore this concept further:
1. How does the angle of the applied force affect the normal force and friction?
2. What happens to the normal force if the angle increases further?
3. How would the work done change if there were no friction?
4. Can you calculate the power generated if the cart was pushed for a certain amount of time?
5. How does the coefficient of kinetic friction impact the total work done by the shopper?

**Tip**: Always check the direction of forces carefully when analyzing problems involving angles, friction, and inclined planes!

A shopper pushes her 25 kg grocery cart by a force of 2250 N inclined at an angle of 60° with the horizontal through a distance of 7.5 m. Find the work done by (a) the 2250 N force and (b) friction. Assume that the coefficient of kinetic friction is 0.32.

Learn how to calculate the work done by a 2250 N force inclined at 60° and the effect of kinetic friction on a 25 kg grocery cart over a distance of 7.5 meters. This problem includes concepts such as work, forces, friction, and trigonometry.

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