Question

An object of mass 2 Kg is lifted up to height 2 m. The work done will be :

• A. \(39.20 \ \text{J}\)
• B. \(9.80 \ \text{J}\)
• C. \(98 \ \text{J}\)
• D. \(980 \ \text{J}\)

Hint:

When calculating work done against gravity: - Use the formula \(W = mgh\). - Remember that \(g\) is the acceleration due to gravity, approximately \(9.8 \ \text{m/s}^2\) on Earth. Sometimes, for simpler calculations, \(g = 10 \ \text{m/s}^2\) might be used if specified or if options suggest it, but \(9.8 \ \text{m/s}^2\) is more precise. - Ensure all units are in the SI system (mass in Kg, height in m, \(g\) in \(\text{m/s}^2\)) to get the work done in Joules (J). This work done is stored as gravitational potential energy in the object.

Answer 1

The Correct Option is A

Concept: When an object is lifted against gravity, the work done on the object is stored as gravitational potential energy. The work done in lifting an object is equal to the change in its potential energy. Step 1: Identify the given quantities Mass of the object, \(m = 2 \ \text{Kg}\) Height to which the object is lifted, \(h = 2 \ \text{m}\) Acceleration due to gravity, \(g\). The standard value for \(g\) is approximately \(9.8 \ \text{m/s}^2\). We will use this value. Step 2: Recall the formula for work done against gravity (or change in potential energy) The work done (\(W\)) in lifting an object of mass \(m\) to a height \(h\) against gravity is given by: \[ W = mgh \] This formula is derived from \(W = Fd\cos\theta\). Here, the force \(F\) required to lift the object is equal to its weight (\(mg\)), the displacement \(d\) is the height \(h\), and the angle \(\theta\) between the upward lifting force and the upward displacement is \(0^\circ\) (\(\cos 0^\circ = 1\)). Step 3: Substitute the values into the formula \[ W = (2 \ \text{Kg}) \times (9.8 \ \text{m/s}^2) \times (2 \ \text{m}) \] Step 4: Calculate the work done First, multiply the mass and height: \(2 \ \text{Kg} \times 2 \ \text{m} = 4 \ \text{Kg} \cdot \text{m}\) Now, multiply this by \(g\): \[ W = (4 \ \text{Kg} \cdot \text{m}) \times (9.8 \ \text{m/s}^2) \] \[ W = 39.2 \ \text{Kg} \cdot \text{m}^2/\text{s}^2 \] The unit \(\text{Kg} \cdot \text{m}^2/\text{s}^2\) is equivalent to Joules (J), which is the SI unit of work and energy. \[ W = 39.2 \ \text{J} \] Looking at the options, \(39.20 \ \text{J}\) matches our calculation. Therefore, the work done will be \(39.20 \ \text{J}\).