What do you mean by zero work, positive work and negative work?

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In physics, work is defined as the energy transferred to or from an object by means of a force acting on the object. It is expressed mathematically as the product of the force and the displacement of the object in the direction of the force.

  • When the force acting on an object is perpendicular to the displacement of the object, the work done by the force is zero. In other words, no energy is transferred to or from the object in this case.
  • If the force and the displacement of the object are in the same direction, the work done by the force is positive. This means that energy is transferred to the object by the force, and the object gains energy.
  • If the force and the displacement of the object are in opposite directions, the work done by the force is negative. This means that energy is transferred from the object to the force, and the object loses energy.

In summary, zero work means that no energy is transferred to or from the object, positive work means that energy is transferred to the object, and negative work means that energy is transferred from the object.

zero work, positive work and negative work

Zero Work, Positive Work and Negative Work

Previous Year Questions

  1. Calculate the work done by the applied force. [DUET 2011]
  2. What is the work done to pull the hanging part on the table. [JIPMER 2000]
  3. What is the work done in taking charge Q once along the loop? [NEET 2005]
  4. The work done by a given force only depends on? [JKCET 2018]
  5. Find the unknown angle between the force and displacement vector. [AMUEEE 2011]

Also check:

CBSE CLASS XII Related Questions

  • 1.

    A magnet held vertically, with its north pole down, is dropped along the axis of a closed solenoid placed vertically on a table. If the observer looks down from the top,

      • the induced current will flow in the anticlockwise direction.
      • the induced current will flow in the clockwise direction.
      • no induced current will flow in the solenoid.
      • the magnet will fall with a constant velocity.

    • 2.
      Two point charges -Q and Q are located at points (d, 0) and (0, d) respectively, in x-y plane. The electric field E at the origin will be:

        • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (\hat{i} - \hat{j})\)
        • \(\frac{1}{4\pi\epsilon_0} \frac{\sqrt{2}Q}{d^2} (-\hat{i} - \hat{j})\)
        • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (\hat{i} - \hat{j})\)
        • \(\frac{1}{4\pi\epsilon_0} \frac{Q}{d^2} (-\hat{i} - \hat{j})\)

      • 3.
        In a region, the electric potential varies as \(V = 10 - 50x\), where \(V\) is in volts and \(x\) is in meters. The electric field in the region is

          • \(10\ \text{N/C}\) along \(+x\)
          • \(10\ \text{N/C}\) along \(-x\)
          • \(50\ \text{N/C}\) along \(+x\)
          • \(50\ \text{N/C}\) along \(-x\)

        • 4.

          A rectangular loop of size 5 cm $\times$ 8 cm is lying in the $x$-$y$ plane in a uniform magnetic field $\vec{B} = (2.0\ \text{T})\hat{k}$. The total magnetic flux linked with the loop is:

            • 80 Wb
            • 16 Wb
            • \(8 \times 10^{-2}\) Wb
            • \(8 \times 10^{-3}\) Wb

          • 5.

            A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is (i) $\sqrt{2}$, and (ii) $\sqrt{3}$.


              • 6.
                A wire of length \(L\) is bent round into (i) a square coil having \(N\) turns and (ii) a circular coil having \(N\) turns. The coil in both cases is free to turn about a vertical axis coinciding with the plane of the coil, in a uniform, horizontal magnetic field and carry the same currents. Find the ratio of the maximum value of the torque acting on the square coil to that on the circular coil.

                  CBSE CLASS XII Previous Year Papers

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