Final answer:
The decrease in gravitational potential energy of the lorry is calculated to be 98,000 Joules, and its speed at the bottom of the slope is approximately 6.26 m/s, based on the conservation of energy.
Step-by-step explanation:
To calculate the decrease in gravitational potential energy (GPE) of a 5000 kg lorry parked on a 20m long 1 in 10 slope, we need to first determine the height (h) of the slope. Because the slope is 1 in 10, for every 10m along the slope, the vertical height changes by 1m. For a 20m slope, this gives us a height of 2m. The formula for GPE is Mgh, where M is mass, g is the acceleration due to gravity (9.8 m/s2), and h is the height. Plugging in the values:
GPE = 5000 kg × 9.8 m/s2 × 2 m = 98,000 J
To calculate the speed at the bottom of the slope, we apply the conservation of energy principle where the initial potential energy is converted into kinetic energy (KE). Since all the potential energy is converted into kinetic energy (ignoring air resistance and friction), GPE is equal to KE. The formula for KE is (1/2)mv2. Setting GPE equal to KE gives us:
98,000 J = (1/2) × 5000 kg × v2
Solving for v, we get:
v = sqrt((2 × 98,000 J) / 5000 kg) = sqrt(39.2) = 6.26 m/s (approximately)
Therefore, the speed of the lorry at the bottom of the slope is approximately 6.26 m/s.