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Grade 9Electric Current

A horse pulls a wagon of mass 1000kg from rest against a constant resistance of 100Nalong a horizontal floor. The horizontal pull exerted initially is 600N and decreases uniformly with the distance covered to 400N at a distance of 15m from start.The velocity of wagon at this point is n*(3)^1/2.find n

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12 Years agoGrade 9
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ApprovedApproved Tutor Answer1 Year ago

To solve this problem, we need to analyze the forces acting on the wagon and apply the principles of physics, particularly Newton's laws of motion and the work-energy principle. Let's break it down step by step.

Understanding the Forces

The wagon has a mass of 1000 kg and is being pulled by a horse. Initially, the horse exerts a force of 600 N, but this force decreases uniformly to 400 N over a distance of 15 m. Meanwhile, there is a constant resistance force of 100 N acting against the motion of the wagon.

Net Force Calculation

The net force acting on the wagon at any point can be calculated by subtracting the resistance from the pulling force. Let's denote the pulling force as F:

  • At the start (0 m): F = 600 N
  • At 15 m: F = 400 N

Thus, the net force at the start is:

Net Force (0 m) = 600 N - 100 N = 500 N

And at 15 m:

Net Force (15 m) = 400 N - 100 N = 300 N

Applying Newton's Second Law

According to Newton's second law, the acceleration (a) can be calculated using the formula:

F = m * a

Where:

  • F is the net force
  • m is the mass (1000 kg)
  • a is the acceleration

Acceleration Calculation

At the start (0 m):

500 N = 1000 kg * a

Solving for a gives:

a = 500 N / 1000 kg = 0.5 m/s²

Now, at 15 m:

300 N = 1000 kg * a

Solving for a gives:

a = 300 N / 1000 kg = 0.3 m/s²

Finding the Average Acceleration

Since the acceleration changes uniformly from 0.5 m/s² to 0.3 m/s² over the distance of 15 m, we can find the average acceleration (a_avg):

a_avg = (0.5 m/s² + 0.3 m/s²) / 2 = 0.4 m/s²

Using Kinematic Equations

To find the final velocity (v) of the wagon after traveling 15 m, we can use the kinematic equation:

v² = u² + 2 * a * s

Where:

  • u is the initial velocity (0 m/s, since it starts from rest)
  • a is the average acceleration (0.4 m/s²)
  • s is the distance (15 m)

Substituting the values:

v² = 0 + 2 * 0.4 m/s² * 15 m

v² = 12 m²/s²

v = √12 m/s = 2√3 m/s

Finding the Value of n

According to the problem, the velocity at this point is given as n * (3)^(1/2). We have found that:

v = 2√3 m/s

Setting these equal gives:

n * √3 = 2√3

Dividing both sides by √3:

n = 2

Thus, the value of n is 2.