In a game called 'High Striker' a person hits a button with a hammer - Edexcel - A-Level Physics - Question 17 - 2023 - Paper 1

To analyze if the cylinder hits the bell, we first calculate the work done on the hammer head:

$ext{Work (W)} = ext{Force (F)} imes ext{Distance (d)}$

Here, the force exerted by the hammer head is 58 N, and the distance is 1.2 m:

$W = 58 ext{ N} imes 1.2 ext{ m} = 69.6 ext{ J}$

Next, we apply the efficiency of energy transfer:

$ext{Useful Energy Output} = ext{Efficiency} imes ext{Total Energy Input}$

Given the efficiency is 4.0%, the useful energy output is:

$ext{Useful Energy Output} = 0.04 imes 69.6 ext{ J} = 2.784 ext{ J}$

Next, we calculate the gravitational potential energy gained by the cylinder when it moves upward:

$ext{Potential Energy (PE)} = mgh$

Where:

• m = 0.15 kg (mass of the cylinder)
• g = 9.81 m/s² (acceleration due to gravity)
• h = 2.7 m (height to hit the bell)

Thus,

$PE = 0.15 ext{ kg} imes 9.81 ext{ m/s}^2 imes 2.7 ext{ m} = 0.15 imes 9.81 imes 2.7 ext{ J} = 3.98 ext{ J}$

Comparing the useful energy output and potential energy: 2.784 J < 3.98 J. Therefore, the energy provided is insufficient for the cylinder to hit the bell.

When the velocity of the hammer head doubles, the kinetic energy, which is dependent on the square of the velocity, increases by a factor of four, as represented by the equation:

KE = rac{1}{2}mv^2

However, due to the energy losses inherent in any system (in this case, the efficiency of energy transfer), only a portion of this increased energy will lead to an increase in height. Thus, although more energy is imparted to the cylinder, the energy converted into gravitational potential energy does not translate directly to double the height gained, due to inefficiencies in the conversion process.