A 12.7 kilogram package is pulled with a 28.5 newton force at a 60.3 degree angle, across the ground where the coefficient of friction is equal to 0.0802. What is the force of friction on the package?
Force equals blank newtons.
Remember to indicate if the force is positive with a plus sign or negative with a minus sign.

To find the force of friction on the package, we need to calculate the frictional force using the formula:

\[
f_{\text{friction}} = \mu \cdot N
\]

Where:
- \( \mu \) is the coefficient of friction (0.0802),
- \( N \) is the normal force.

The normal force \( N \) can be calculated by considering the forces acting vertically on the package. The weight of the package and the vertical component of the applied force affect the normal force. Here's the process:

1. **Weight of the package**: 
   \[
   W = m \cdot g = 12.7 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 124.587 \, \text{N}
   \]

2. **Vertical component of the applied force**:
   The vertical component of the applied force \( F_y \) is:
   \[
   F_y = F \cdot \sin(\theta) = 28.5 \, \text{N} \cdot \sin(60.3^\circ)
   \]

3. **Normal force**:
   The normal force \( N \) is the weight minus the vertical component of the applied force:
   \[
   N = W - F_y
   \]

4. **Frictional force**:
   Once we have \( N \), the frictional force can be found using \( f_{\text{friction}} = \mu \cdot N \).

Let me calculate these values.The force of friction on the package is \( +8.01 \, \text{N} \).

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How would the force of friction change if the coefficient of friction was higher?
2. What would happen if the angle of the applied force increased to 90 degrees?
3. How can you calculate the horizontal component of the applied force?
4. What is the net force acting on the package?
5. How would the acceleration of the package be determined from the net force?

**Tip**: Friction always opposes the motion or attempted motion between two surfaces in contact.

A 12.7 kilogram package is pulled with a 28.5 newton force at a 60.3 degree angle, across the ground where the coefficient of friction is equal to 0.0802. What is the force of friction on the package?

Learn how to calculate the frictional force on a 12.7 kg package pulled with a 28.5 N force at a 60.3° angle. This problem involves key physics concepts such as friction, normal force, and Newton's Second Law. Suitable for high school students in Grades 10-12.

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