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A block of mass 1 kg is pushed up a surface inclined to the horizontal at an angle of 30 degrees by a force of 10 N parallel to the inclined surface. The coefficient of friction between the block and inclination is 0.1. If the block is pushed up by 10 m along the inclination, calculate:

A) Work done against gravity
B) Work done against the force of friction
C) Increase in potential energy
D) Increase in kinetic energy
E) Work done by the applied force.

1 Answer

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Final answer:

The work done against gravity is 49 J, work done against the force of friction is 8.5 J, the increase in potential energy is 49 J, the increase in kinetic energy is 42.5 J, and the work done by the applied force is 100 J.

Step-by-step explanation:

To solve these problems, we need to apply the principles of work and energy. The work done by a force is defined as the product of the force and the displacement in the direction of the force:

Work = Force × Displacement × cos(θ)

Where θ is the angle between the force and the displacement.

  1. Work done against gravity: This is equal to the weight of the block (mass × acceleration due to gravity) multiplied by the vertical displacement. The weight of the block is 1 kg × 9.8 m/s² = 9.8 N. The vertical displacement is 10 m × sin(30°) = 5 m. Therefore, the work done against gravity is Work = 9.8 N × 5 m = 49 J.
  2. Work done against the force of friction: The force of friction is given by the normal force (which is the weight of the block times cos(30°)) times the coefficient of friction. The work done by friction is then the force of friction times displacement along the incline. The work done against friction is thus Work = 1 kg × 9.8 m/s² × cos(30°) × 0.1 × 10 m = 8.5 J.
  3. Increase in potential energy: The increase in potential energy is the same as the work done against gravity, which is 49 J.
  4. Increase in kinetic energy: This is equal to the work done by the applied force minus the sum of the work done against gravity and the work done against friction: Increase in kinetic energy = 10 N × 10 m - (49 J + 8.5 J) = 42.5 J.
  5. Work done by the applied force: The work done by the applied force is the force times the displacement along the incline, which is Work = 10 N × 10 m = 100 J.

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