Work, Energy and Power MCQ question and answer with solution | Physics Gk MCQs

Understanding the Question:
This question is about work, energy, and power. It asks us to find the force needed to stop a moving car.

What we know:
* The car's initial speed is 72 km/h. We need to convert this to meters per second (m/s) because that's the standard unit for physics calculations. There are 1000 meters in a kilometer and 3600 seconds in an hour, so 72 km/h = (72 * 1000) / 3600 = 20 m/s.
* The car stops in 4 seconds (this is the time it takes for the car to decelerate to 0 m/s).
* The car's mass is 1000 kg.

What we need to find:
We need to calculate the force applied by the brakes to bring the car to a complete stop.

How to solve it:
First, we need to find the deceleration (or negative acceleration) of the car. Deceleration is the change in speed divided by the time it takes for that change to occur. The change in speed is from 20 m/s to 0 m/s, and the time is 4 seconds.
Deceleration = (final speed - initial speed) / time = (0 - 20 m/s) / 4 s = -5 m/s² (The negative sign indicates deceleration).
Next, we use Newton's second law of motion: Force = mass × acceleration. In this case, the acceleration is actually a deceleration, so we use the negative value.
Force = 1000 kg × (-5 m/s²) = -5000 N
The magnitude of the force (ignoring the negative sign, which just tells us the direction) is 5000 N, which is the same as 5.0 × 10³ N.

Therefore, the correct answer is D: 5.0 × 10³ N

To find the height, we use the formula for gravitational potential energy:

Potential Energy (PE) = m × g × h

Where:

PE = 9800 J (Energy used)

m = 70 kg (Mass of the object)

g = 10 m/s² (Acceleration due to gravity, commonly used in numerical problems)

h = ? (Height)

Substituting the given values:

9800 = 70 × 10 × h

9800 = 700h

h = 9800 ÷ 700 = 14 m

Therefore, the 70 kg object was lifted to a height of 14 metres.

Note: Options with negative height are incorrect because the object is being lifted upward, so the height must be positive.