We read your photo as:

A 50 kg bag is moved from point A to point B, 10 m away horizontally. An applied force of 250 N is used to move the bag. The vertical height difference between A and B is 4 m. Calculate: i) The work d

Physics

ScanSolve AI Answer

Step-by-step solution

12 steps

Part a(i)· 2 steps

The work done in moving the bag from A to B

  1. Identify the given values for applied force and distance.

    Applied force F=250 NF = 250 \text{ N}
    Distance d=10 md = 10 \text{ m}

  2. Calculate the work done using the formula W=F×dW = F \times d.

    Win=250 N×10 m=2500 JW_{in} = 250 \text{ N} \times 10 \text{ m} = 2500 \text{ J}

    The work done in moving the bag from A to B is 2500 J\boxed{\text{2500 J}}.

Part a(ii)· 2 steps

The useful work done

  1. Identify the given values for mass, height, and acceleration due to gravity.

    Mass m=50 kgm = 50 \text{ kg}
    Height h=4 mh = 4 \text{ m}
    Acceleration due to gravity g=9.8 m/s2g = 9.8 \text{ m/s}^2

  2. Calculate the useful work done (potential energy gained) using the formula W=mghW = mgh.

    Wout=50 kg×9.8 m/s2×4 m=1960 JW_{out} = 50 \text{ kg} \times 9.8 \text{ m/s}^2 \times 4 \text{ m} = 1960 \text{ J}

    The useful work done is 1960 J\boxed{\text{1960 J}}.

Part a(iii)· 2 steps

What is the efficiency of the arrangement?

  1. Use the calculated work input (WinW_{in}) and useful work output (WoutW_{out}).

    Win=2500 JW_{in} = 2500 \text{ J}
    Wout=1960 JW_{out} = 1960 \text{ J}

  2. Calculate the efficiency using the formula η=WoutWin×100%\eta = \frac{W_{out}}{W_{in}} \times 100\%.

    η=1960 J2500 J×100%=0.784×100%=78.4%\eta = \frac{1960 \text{ J}}{2500 \text{ J}} \times 100\% = 0.784 \times 100\% = 78.4\%

    The efficiency of the arrangement is 78.4%\boxed{\text{78.4\%}}.

Part a(iv)

Give one reason why the efficiency is less than 100%

Efficiency is less than 100% due to friction between the bag and the inclined plane, which converts some of the input energy into heat.

Part b(i)· 4 steps

The change in the kinetic energy

  1. Identify the given values for mass and speeds.

    Mass m=1000 kgm = 1000 \text{ kg}
    Initial speed u=20 m/su = 20 \text{ m/s}
    Final speed v=3×20 m/s=60 m/sv = 3 \times 20 \text{ m/s} = 60 \text{ m/s}

  2. Calculate the initial kinetic energy (KEiKE_i).

    KEi=12mu2=12×1000 kg×(20 m/s)2=500 kg×400 m2/s2=200000 JKE_i = \frac{1}{2}mu^2 = \frac{1}{2} \times 1000 \text{ kg} \times (20 \text{ m/s})^2 = 500 \text{ kg} \times 400 \text{ m}^2/\text{s}^2 = 200000 \text{ J}

  3. Calculate the final kinetic energy (KEfKE_f).

    KEf=12mv2=12×1000 kg×(60 m/s)2=500 kg×3600 m2/s2=1800000 JKE_f = \frac{1}{2}mv^2 = \frac{1}{2} \times 1000 \text{ kg} \times (60 \text{ m/s})^2 = 500 \text{ kg} \times 3600 \text{ m}^2/\text{s}^2 = 1800000 \text{ J}

  4. Calculate the change in kinetic energy (ΔKE\Delta KE).

    ΔKE=KEf−KEi=1800000 J−200000 J=1600000 J\Delta KE = KE_f - KE_i = 1800000 \text{ J} - 200000 \text{ J} = 1600000 \text{ J}

    The change in the kinetic energy is 1600000 J\boxed{\text{1600000 J}}.

Part b(ii)· 2 steps

The power developed

  1. Use the calculated change in kinetic energy and the given time.

    ΔKE=1600000 J\Delta KE = 1600000 \text{ J}
    Time t=14.5 st = 14.5 \text{ s}

  2. Calculate the power developed using the formula P=ΔKEtP = \frac{\Delta KE}{t}.

    P=1600000 J14.5 s≈110344.83 WP = \frac{1600000 \text{ J}}{14.5 \text{ s}} \approx 110344.83 \text{ W}

    The power developed is approximately 110344.83 W\boxed{\text{110344.83 W}}.

Part c

Alternative energy sources to fossil fuels include sun, wind. For each source, give an example of how energy may be obtained from it.

  • Sun: Energy is obtained from the sun by using solar panels (photovoltaic cells) to convert sunlight directly into electricity, or by using solar thermal collectors to heat water.
  • Wind: Energy is obtained from wind by using wind turbines to convert the kinetic energy of moving air into rotational energy, which then drives a generator to produce electricity.

Part d

Identify the safety symbols A and B in figure 6.

Figure 6 is not provided in the image.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Physics Questions

QA 40 kg box is pulled up an incline of 15 degrees by a horizontal force of 200 N. If the box moves…QA Man need to carry 10 bags of Cement for along a distance of 10m. Each bag of Cement is 10kg…QA box of mass 40 kg is dragged along the floor by a rope inclined at 60° to the horizontal. The…QA 100 N force moves an object 5 m in its direction. How much work is done?QA person carries a bucket of water of mass 20 kg for a distance of 10 m. Calculate the work done.
Browse by topic:Mechanics
See all Physics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp