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A block of mass 2 kg is initially at rest on a horizontal surface. A horizontal force with a constant...

A block of mass 2 kg is initially at rest on a horizontal surface. A horizontal force with a constant magnitude of 8 N is then exerted on the block, and the block remains at rest. The coefficient of static friction between the block and the surface is 0.5.

What is the magnitude of the frictional force exerted on the block by the surface?

Answer

Detailed Answer

To determine the magnitude of the frictional force exerted on the block by the surface, we start with the following information:

  • Mass of the block (m) = 2 kg
  • Applied horizontal force (F_applied) = 8 N
  • Coefficient of static friction (μ_s) = 0.5

Step 1: Calculate the maximum static friction force

The maximum static friction force (F_friction_max) can be calculated using the formula:

F_friction_max = μ_s × N

Where N is the normal force. For an object resting on a horizontal surface without any vertical forces acting on it other than its weight, the normal force (N) is equal to the weight of the block:

N = m × g

Assuming acceleration due to gravity (g) is approximately 9.81 m/s², we calculate N:

N = 2 kg × 9.81 m/s² = 19.62 N

Now, we substitute the value of N into the static friction formula:

F_friction_max = 0.5 × 19.62 N = 9.81 N

Step 2: Determine the actual frictional force

The block is initially at rest, and the applied force of 8 N does not exceed the maximum static friction force of 9.81 N. Hence, the static frictional force will match the applied force to keep the block in equilibrium.

Therefore, the actual frictional force (F_friction) is equal to the applied force:

F_friction = F_applied = 8 N

Conclusion

The magnitude of the frictional force exerted on the block by the surface is:

F_friction = 8 N

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