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A 2 kg block slides up a 10 m long, 5 m high smooth incline. If it starts and ends at rest, what is the work done by non-conservative forces?

Physics

ScanSolve AI Answer

Answer

98 J\text{98 J}

Step-by-step solution

5 steps

To find the work done by the non-conservative force, we can use the Work-Energy Theorem, which states that the work done by non-conservative forces (WncW_{nc}) is equal to the change in the total mechanical energy (ΔEmech\Delta E_{mech}). The change in mechanical energy is the sum of the change in kinetic energy (ΔKE\Delta KE) and the change in gravitational potential energy (ΔPEg\Delta PE_g).

Since the problem states the incline is "smooth," there is no friction. The non-conservative force FncF_{nc} is the only non-conservative force acting. The problem does not specify any initial or final velocity, so we assume the block starts from rest at the bottom and ends at rest at the top, meaning the change in kinetic energy is zero (ΔKE=0\Delta KE = 0).

  1. Identify the given values.

    Mass of the block, m=2 kgm = 2 \text{ kg}
    Height of the incline, h=5 mh = 5 \text{ m}
    Acceleration due to gravity, g=9.8 m/s2g = 9.8 \text{ m/s}^2 (standard value)
    Length of the incline, d=10 md = 10 \text{ m} (This value is not directly needed for this approach, but it's consistent with the height.)

  2. State the Work-Energy Theorem for non-conservative forces.

    Wnc=ΔEmech=ΔKE+ΔPEgW_{nc} = \Delta E_{mech} = \Delta KE + \Delta PE_g
  3. Calculate the change in kinetic energy (ΔKE\Delta KE).

    Assuming the block starts and ends at rest:

    vi=0 m/sv_i = 0 \text{ m/s} vf=0 m/sv_f = 0 \text{ m/s} ΔKE=12mvf2−12mvi2=12m(0)2−12m(0)2=0 J\Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = \frac{1}{2}m(0)^2 - \frac{1}{2}m(0)^2 = 0 \text{ J}
  4. Calculate the change in gravitational potential energy (ΔPEg\Delta PE_g).

    The block is pulled up from a height of 0 m0 \text{ m} to 5 m5 \text{ m}.

    ΔPEg=mghf−mghi=mg(h)−mg(0)\Delta PE_g = mgh_f - mgh_i = mg(h) - mg(0) ΔPEg=mgh\Delta PE_g = mgh ΔPEg=(2 kg)(9.8 m/s2)(5 m)\Delta PE_g = (2 \text{ kg})(9.8 \text{ m/s}^2)(5 \text{ m}) ΔPEg=98 J\Delta PE_g = 98 \text{ J}
  5. Calculate the work done by the non-conservative force (WncW_{nc}).

    Wnc=ΔKE+ΔPEgW_{nc} = \Delta KE + \Delta PE_g Wnc=0 J+98 JW_{nc} = 0 \text{ J} + 98 \text{ J} Wnc=98 JW_{nc} = 98 \text{ J}

    The work done by the non-conservative force is 98 J\text{98 J}.

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