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Grade 12General Physics

Consider the inclined plane with an angle of 67 degree. The spring constant has a value of 1850 N/m. A 10kg block is pushed against the spring compressing it to 40 cm. The block is then released and the spring expands pushing the block up the plane. the coefficient of kinetic friction between the block and the plane is 0.5. what is the normal force and frictional force acting on the system? how far up the plane will the block slide before coming to stop? what is the change in KE

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8 Years agoGrade 12
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ApprovedApproved Tutor Answer1 Year ago

To tackle this problem, we need to break it down into several parts: calculating the normal force, the frictional force, the distance the block will slide up the inclined plane, and the change in kinetic energy (KE). Let's go through each step methodically.

Calculating the Normal Force

The normal force is the force exerted by a surface that supports the weight of an object resting on it. On an inclined plane, the normal force can be calculated using the formula:

Normal Force (N) = m * g * cos(θ)

Where:

  • m = mass of the block = 10 kg
  • g = acceleration due to gravity ≈ 9.81 m/s²
  • θ = angle of the incline = 67 degrees

First, we need to convert the angle from degrees to radians if necessary, but for this calculation, we can use the cosine directly:

N = 10 kg * 9.81 m/s² * cos(67°)

Calculating this gives:

N ≈ 10 * 9.81 * 0.3907 ≈ 38.34 N

Determining the Frictional Force

The frictional force can be calculated using the formula:

Frictional Force (f) = μ * N

Where:

  • μ = coefficient of kinetic friction = 0.5
  • N = normal force calculated previously ≈ 38.34 N

Now, substituting the values:

f = 0.5 * 38.34 N ≈ 19.17 N

Distance the Block Slides Up the Plane

To find out how far the block will slide up the incline before coming to a stop, we first need to determine the initial energy provided by the spring when it is released. The potential energy stored in the spring can be calculated using:

Potential Energy (PE) = 0.5 * k * x²

Where:

  • k = spring constant = 1850 N/m
  • x = compression of the spring = 0.4 m

Calculating the potential energy:

PE = 0.5 * 1850 N/m * (0.4 m)² ≈ 0.5 * 1850 * 0.16 ≈ 148 N·m

As the block moves up the incline, this energy is converted into work done against friction and gravitational potential energy. The work done against friction can be expressed as:

Work done against friction = f * d

Where d is the distance the block travels up the incline. The gravitational potential energy gained by the block can be expressed as:

Potential Energy Gain = m * g * h

Since h = d * sin(θ), we can rewrite the potential energy gain as:

Potential Energy Gain = m * g * (d * sin(θ))

Setting the initial potential energy equal to the work done against friction plus the potential energy gain gives us:

PE = f * d + m * g * (d * sin(θ))

Substituting the known values:

148 N·m = 19.17 N * d + 10 kg * 9.81 m/s² * (d * sin(67°))

Calculating the sine component:

sin(67°) ≈ 0.9205

Now we can substitute this into our equation:

148 = 19.17d + 10 * 9.81 * (d * 0.9205)

148 = 19.17d + 90.82d

148 = 109.99d

Solving for d gives:

d ≈ 1.35 m

Change in Kinetic Energy

The change in kinetic energy can be determined by considering the initial kinetic energy when the block is released from the spring and the final kinetic energy when it comes to a stop. Initially, the kinetic energy is equal to the potential energy stored in the spring:

Change in KE = Initial KE - Final KE

Since the block comes to a stop, the final kinetic energy is 0. Thus:

Change in KE = PE = 148 N·m

In summary, we have:

  • Normal Force: ≈ 38.34 N
  • Frictional Force: ≈ 19.17 N
  • Distance the block slides up the plane: ≈ 1.35 m
  • Change in Kinetic Energy: ≈ 148 J