The correct answer is D.

Friction typically does negative work because it opposes motion. However, in some cases, friction can do positive work. An example is when a car accelerates, and the friction between the tires and the road helps the car move forward (I explain this point in lecture video, because this point need video explanation)

The frictional force pushes the car in the direction of motion, aiding its acceleration. Thus, in rolling motion, friction can do positive work by helping the object move.

If there is no relative motion between surfaces, then no friction is acting, and thus, the work done by friction is zero

The correct answer is A.

• Northward displacement = 10 meters
• Eastward displacement = 20 meters

r= 20i 10j

The displacement (d) is the hypotenuse of a right triangle with sides 10 meters and 20 meters. Using the Pythagorean theorem:

d = √(10² 20²)

= √(100 400)

= √500 ≈ 22.36 m

The correct answer is A.

W = F . d

The net force is calculated using Newton’s second law:

F = m . a
F = 2 kg × 5 m/s² = 10 N

W = 10 N × 2 m = 20 J

So, the net work done on the object is 20 J.

The correct answer is A.

To calculate the net displacement, we need to consider all the movements in each direction (north, south, and upward).

1. The airplane moves 10 meters north.
2. Then, it moves 6 meters south.

The net displacement in the vertical (north-south) direction is:

Net vertical displacement = 10 m north − 6 m south = 4 m north

Finally, the airplane ascends 3 meters vertically.

The total vertical displacement is:

Total displacement = √(4² + 3²)

= √(16 + 9) = √(25)

= 5 m

The correct answer is D.

W = F . d . cos(θ)

The maximum work occurs when the angle θ = 0°, because cos(0°) = 1, W = W_max.

Since W_max = F . d, we can solve for θ by:

F . d = F . d . cos(θ)

1 = cos(θ)

Taking the inverse cosine of both sides:

θ = cos⁻¹(1/2) = 60°

So, the angle at which the work done will be half of its maximum value is 60°.