Math Problem Statement

A 1.5 kg block is held at rest near the top of a rough incline that makes an angle of 37° from the horizontal. The block is then released and starts moving with a constant acceleration. When the block travels 2.0 m, its velocity is 4.0 m/s. The coefficient of kinetic friction between the surface of the incline and the block is most nearly?

Solution

The problem asks to find the coefficient of kinetic friction (μk\mu_k) between a block and an inclined surface.

**Given dataThe problem asks to find the coefficient of kinetic friction (μk\mu_k) between a block and an inclined surface.

**Given **

  • Mass of the block: m=1.5kgm = 1.5 \, \text{kg}
  • Angle of the incline: θ=37\theta = 37^\circ
  • Distance traveled: d=2.0md = 2.0 \, \text{m}
  • Final velocity: v=4.0m/sv = 4.0 \, \text{m/s}
  • Initial velocity: u=0m/su = 0 \, \text{m/s} (since the block is released from rest)
  • Acceleration is constant.

Step 1: Use kinematic equation to find acceleration (aa)

We can use the equation: v2=u2+2adv^2 = u^2 + 2ad Substituting the known values: (4.0m/s)2=(0m/s)2+2a(2.0m)(4.0 \, \text{m/s})^2 = (0 \, \text{m/s})^2 + 2a(2.0 \, \text{m}) 16=4a16 = 4a a=4m/s2a = 4 \, \text{m/s}^2

Step 2: Analyze forces along the incline

The forces acting on the block along the incline are:

  • Gravitational force component along the incline: Fg=mgsinθF_g = mg \sin \theta
  • Kinetic friction force: Ff=μkmgcosθF_f = \mu_k mg \cos \theta

Using Newton's second law (F=maF = ma), the net force along the incline is: mgsinθμkmgcosθ=mamg \sin \theta - \mu_k mg \cos \theta = ma

Step 3: Solve for the coefficient of kinetic friction (μk\mu_k)

Simplify the equation: gsinθμkgcosθ=ag \sin \theta - \mu_k g \cos \theta = a Substitute g=9.8m/s2g = 9.8 \, \text{m/s}^2, a=4m/s2a = 4 \, \text{m/s}^2, and θ=37\theta = 37^\circ: 9.8sin37μk9.8cos37=49.8 \sin 37^\circ - \mu_k 9.8 \cos 37^\circ = 4 Using the values sin370.6018\sin 37^\circ \approx 0.6018 and cos370.7986\cos 37^\circ \approx 0.7986, we get: 9.8(0.6018)μk9.8(0.7986)=49.8(0.6018) - \mu_k 9.8(0.7986) = 4 5.89767.827μk=45.8976 - 7.827 \mu_k = 4 5.89764=7.827μk5.8976 - 4 = 7.827 \mu_k 1.8976=7.827μk1.8976 = 7.827 \mu_k μk=1.89767.8270.2425\mu_k = \frac{1.8976}{7.827} \approx 0.2425

Step 4: Select the closest answer

The closest answer to μk=0.2425\mu_k = 0.2425 is (B) 0.25.


Would you like further clarification on any step?

Here are 5 related questions:

  1. How does the angle of incline affect the motion of the block?
  2. What would happen if the surface had no friction?
  3. How is the normal force on the block related to the incline's angle?
  4. What is the significance of constant acceleration in this problem?
  5. How can we derive the value of kinetic friction in real-world problems?

Tip: Always double-check the kinematic equations for accuracy when solving physics problems involving motion.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Forces
Friction

Formulas

v^2 = u^2 + 2ad
F = ma
Frictional force (F_f) = μ_k * Normal force
Gravitational force component: F_g = mg * sin(θ)

Theorems

Newton's Second Law

Suitable Grade Level

Grades 11-12