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

To calculate the change in gravitational potential energy, we use the formula:

$E_{p} = mgh$

Assuming the mass of the student is 50 kg and the height is 20.0 m:

$E_{p} = 50 imes 9.8 imes 20 = 9800 ext{ J}$

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

Since 80% of the change in gravitational potential energy has been transferred to the student's kinetic energy store:

$ext{Kinetic Energy gained} = 0.80 imes E_{p} = 0.80 imes 9800 = 7840 ext{ J}$

Therefore, the student's kinetic energy store has increased by 7840 J.

To find the speed ($v$) of the student after falling 20.0 m, we use the kinetic energy equation:

$KE = \frac{1}{2}mv^2$

Substituting in the kinetic energy gained:

$7840 = \frac{1}{2} \times 50 \times v^2$

Rearranging gives:

$v^2 = \frac{7840}{25} = 313.6$

Thus,

$v = \sqrt{313.6} \approx 17.7 ext{ m/s}$

Rounding this to two significant figures gives:

$v = 18 ext{ m/s}$

The energy stored in the bungee cord at the lowest point of the jump is given as 24.5 kJ, which converts to:

$E = 24500 ext{ J}$

Using the formula for the potential energy stored in a spring:

$E = \frac{1}{2} k x^2$

where:

• $k$ = spring constant (N/m)
• $x$ = extension in meters (m)

From the problem, the extension is given as 35 m. Rearranging the formula to find $k$ gives:

$k = \frac{2E}{x^2} = \frac{2 \times 24500}{35^2} = 40 ext{ N/m}$

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