1.7.2 Work (3)
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1.
A mechanic uses a lever to lift a car. The force applied to the long end of the lever is 500 N, and the distance from the fulcrum to the point where the car is lifted is 2 m. Calculate the mechanical advantage of the lever and the amount of work done on the car.
Given:
- Force applied (F) = 500 N
- Distance to car (d) = 2 m
Mechanical Advantage (MA): MA = Output Force / Input Force = F / F = 2 m / 500 N = 0.004 m. This is incorrect. The mechanical advantage is the ratio of the distance over which the input force is applied to the distance over which the output force is applied. In this case, the input force is applied over 2m and the output force is applied over a distance of 2m. Therefore, the MA = 2/2 = 1. This means the force is not multiplied. The question is flawed as it does not provide enough information to calculate the work done on the car. Assuming the car is lifted a distance of 0.2m, the work done on the car is:
Work done on the car: W = F d = 500 N × 0.2 m = 100 J
Answer: The mechanical advantage of the lever is 1. The work done on the car is 100 J.
2.
A 10 A current flows through a resistor of 10 ohms. Calculate the electrical power dissipated by the resistor. Explain how this electrical power relates to the energy transferred in a circuit. Include a brief explanation of the relationship between power, energy, and time.
Answer:
- Calculate the electrical power: P = I2R = (10 A)2 × 10 Ω = 100 W.
- The electrical power dissipated by the resistor is equal to the rate at which electrical energy is converted into heat. This is a direct consequence of the fundamental relationship between power, energy, and time: Energy = Power × Time. In a circuit, the power dissipated by a resistor is the electrical energy transferred to the resistor per unit time. The electrical energy is transferred from the source through the circuit and is dissipated as heat in the resistor.
- The electrical power is directly proportional to the energy transferred in a circuit. A higher power means more energy is transferred in a given time. The relationship is clearly expressed in the formula Energy = Power × Time.
3.
A spring is stretched from its equilibrium position by 0.1 m. The spring constant is 500 N/m. Calculate the amount of elastic potential energy stored in the spring.
Given:
- Extension (x) = 0.1 m
- Spring constant (k) = 500 N/m
Equation: Elastic Potential Energy (ΔE) = ½ k x²
Calculation:
ΔE = ½ × 500 N/m × (0.1 m)² = ½ × 500 × 0.01 J = 2.5 J
Answer: The amount of elastic potential energy stored in the spring is 2.5 J.