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Write down the equation which links gravitational field strength, gravitational potential energy, height and mass - AQA - GCSE Physics - Question 11 - 2018 - Paper 1

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Write down the equation which links gravitational field strength, gravitational potential energy, height and mass. Calculate the change in gravitational potential e... show full transcript

Worked Solution & Example Answer:Write down the equation which links gravitational field strength, gravitational potential energy, height and mass - AQA - GCSE Physics - Question 11 - 2018 - Paper 1

Step 1

Write down the equation which links gravitational field strength, gravitational potential energy, height and mass.

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Answer

The equation linking gravitational field strength (g), gravitational potential energy (E_p), height (h), and mass (m) is:

Ep=mghE_p = mgh

Step 2

Calculate the change in gravitational potential energy from the position where the student jumps to the point 20.0 m below.

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Answer

Assuming the mass of the student is 50 kg, the change in gravitational potential energy can be calculated as:

Ep=mgh=50extkgimes9.8extN/kgimes20.0extm=9800extJE_p = mgh = 50 ext{ kg} imes 9.8 ext{ N/kg} imes 20.0 ext{ m} = 9800 ext{ J}

Thus, the change in gravitational potential energy is 9800 J.

Step 3

How much has the student's kinetic energy store increased after falling 20.0 m?

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Answer

80% of the change in gravitational potential energy has been transferred to the kinetic energy store. Therefore:

extKineticEnergygained=0.8imes9800extJ=7840extJ ext{Kinetic Energy gained} = 0.8 imes 9800 ext{ J} = 7840 ext{ J}

Step 4

Calculate the speed of the student after falling 20.0 m.

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Answer

Using the kinetic energy formula:

KE=12mv2KE = \frac{1}{2} mv^2

We can rearrange to find the speed (v):

v=2KEm=27840extJ50extkg17.7extm/sv = \sqrt{\frac{2 \cdot KE}{m}} = \sqrt{\frac{2 \cdot 7840 ext{ J}}{50 ext{ kg}}} \approx 17.7 ext{ m/s}

Therefore, the speed of the student after falling 20.0 m is approximately 18 m/s when rounded to two significant figures.

Step 5

Calculate the spring constant of the bungee cord.

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Answer

The energy stored by the bungee cord is given as 24.5 kJ, which is equivalent to 24500 J. Using the formula for elastic potential energy:

E=12kx2E = \frac{1}{2}kx^2

Where k is the spring constant and x is the extension (35 m):

Rearranging gives:

k=2Ex2=224500extJ352=40extN/mk = \frac{2E}{x^2} = \frac{2 \cdot 24500 ext{ J}}{35^2} = 40 ext{ N/m}

Thus, the spring constant of the bungee cord is 40 N/m.

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