Question

A block of mass 15.0 kg slides down a ramp inclined at 28.0 �� above the horizontal. As it slides, a kinetic friction force of 30.0 N parallel to the ramp acts on it. If the block slides for 7.50 m along the ramp, find the work done on the block by friction.

Answer 1

The coefficient of friction is approximately 0.363, and the block will slide at a constant velocity when the incline angle is approximately 14°.

Step-by-step explanation:

Given:

• Mass of block (m) = 2 kg
• Incline angle (θ) = 25°
• Frictional force (F_friction) = 4.86 N
• Acceleration due to gravity (g) = 10 m/s²

To find:

1. Coefficient of friction (μ)
2. Incline angle for constant velocity

Solution:

1. The coefficient of friction (μ) can be calculated using the formula:

μ = F_friction / (m * g * cos(θ))

Substituting the given values, we get:

μ = 4.86 N / (2 kg * 10 m/s² * cos(25°))

μ ≈ 0.363

Therefore, the coefficient of friction is approximately 0.363.

2. The block will slide at a constant velocity when the force of friction is equal to the force component down the plane. This can be calculated using the formula:

F_friction = m * g * sin(θ)

Substituting the given values, we get:

4.86 N = 2 kg * 10 m/s² * sin(θ)

sin(θ) ≈ 0.243

θ ≈ 14°

Therefore, the block will slide at a constant velocity when the incline angle is approximately 14°.