Math Problem Statement

A block of mass 2.02 kg is accelerated across a rough surface by a rope passing over a pulley. The tension in the rope is 13.8 N, and the pulley is 11.9 cm above the top of the block. The coefficient of kinetic friction is 0.375. Determine the acceleration of the block when x = 0.410 m.

Solution

To solve this problem, we need to calculate the acceleration of the block when x=0.410mx = 0.410 \, \text{m}. Here's a breakdown of the forces involved and how to calculate the acceleration.

Given dataTo solve this problem, we need to calculate the acceleration of the block when x=0.410mx = 0.410 \, \text{m}. Here's a breakdown of the forces involved and how to calculate the acceleration.

Given

  • Mass of the block, m=2.02kgm = 2.02 \, \text{kg}
  • Tension in the rope, T=13.8NT = 13.8 \, \text{N}
  • Coefficient of kinetic friction, μk=0.375\mu_k = 0.375
  • Distance x=0.410mx = 0.410 \, \text{m}
  • Pulley height above the block is h=11.9cm=0.119mh = 11.9 \, \text{cm} = 0.119 \, \text{m}

Step 1: Normal Force and Friction

The block is on a horizontal surface, so the normal force is equal to the weight of the block. The frictional force is given by: fk=μkN=μkmgf_k = \mu_k \cdot N = \mu_k \cdot m \cdot g where g=9.81m/s2g = 9.81 \, \text{m/s}^2 is the acceleration due to gravity.

Step 2: Force Analysis in the Horizontal Direction

We need to find the horizontal component of the tension force TxT_x. The total tension TT acts at an angle because the pulley is elevated by a height hh.

First, we calculate the angle θ\theta from the geometry of the problem: θ=tan1(hx)=tan1(0.1190.410)\theta = \tan^{-1}\left(\frac{h}{x}\right) = \tan^{-1}\left(\frac{0.119}{0.410}\right) Once we have θ\theta, the horizontal component of the tension is: Tx=Tcos(θ)T_x = T \cdot \cos(\theta)

Step 3: Applying Newton's Second Law

The net force in the horizontal direction is the difference between the horizontal component of the tension and the kinetic friction force: Fnet=TxfkF_{\text{net}} = T_x - f_k The acceleration of the block is then: a=Fnetma = \frac{F_{\text{net}}}{m}

Let me calculate the values for the tension component, friction, and acceleration.The acceleration of the block is approximately 2.88m/s22.88 \, \text{m/s}^2, and the angle θ\theta at which the rope pulls the block is about 16.1916.19^\circ.

Would you like further details or explanations on any part of this solution?

Here are some related questions for further exploration:

  1. How would the acceleration change if the coefficient of kinetic friction were higher?
  2. What happens to the acceleration if the mass of the block increases?
  3. How does the angle of the rope affect the horizontal component of the tension force?
  4. What would be the effect of reducing the tension in the rope to 10 N?
  5. How would the acceleration differ if the pulley were at a lower height?

Tip: The angle of a rope in pulley problems is crucial since it changes the effective tension in the horizontal direction. Always resolve forces into components!

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Math Problem Analysis

Mathematical Concepts

Newton's Laws of Motion
Trigonometry
Kinetic Friction
Vector Decomposition

Formulas

f_k = μ_k * N
θ = tan^(-1)(h/x)
T_x = T * cos(θ)
F_net = T_x - f_k
a = F_net/m

Theorems

Newton's Second Law of Motion

Suitable Grade Level

College level Physics or Advanced High School Physics