Question

homework 3
q1 a. define work in physics
b. under what condition is work positive.
c. give one real - life example of positive, negative and zero work

Explanation:

• a. Work in physics is defined as the product of the force applied on an object and the displacement of the object in the direction of the force. Mathematically, \(W = F\times d\times\cos\theta\), where \(F\) is the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and the displacement vectors.
• b. Work is positive when the angle \(\theta\) between the force \(F\) and the displacement \(d\) is between \(0^{\circ}\) and \(90^{\circ}\) (i.e., \(0\leq\theta < 90^{\circ}\)). In this case, \(\cos\theta> 0\), and using the formula \(W = F\times d\times\cos\theta\), the work \(W>0\).
• c.
• Positive work: When a person pushes a box forward on a horizontal surface. The force applied by the person is in the direction of the displacement of the box. Using \(W = F\times d\times\cos\theta\) (where \(\theta = 0^{\circ}\) and \(\cos\theta = 1\)), work done is positive.
• Negative work: When a moving car is stopped by applying brakes. The force of friction acts opposite to the direction of the displacement of the car. Using \(W = F\times d\times\cos\theta\) (where \(\theta= 180^{\circ}\) and \(\cos\theta=- 1\)), work done by the frictional force is negative.
• Zero work: When a person holds a heavy object and walks horizontally. The force applied by the person is in the vertical direction (to counter - act gravity), and the displacement is in the horizontal direction. So, \(\theta = 90^{\circ}\) and \(\cos\theta = 0\). Using \(W = F\times d\times\cos\theta\), the work done \(W = 0\).

Answer:

a. Work in physics is \(W = F\times d\times\cos\theta\) (force - displacement - cosine of the angle between them).
b. Work is positive when \(0\leq\theta < 90^{\circ}\).
c. Positive work: Pushing a box forward; Negative work: Braking a moving car; Zero work: Holding a heavy object while walking horizontally.