Final answer:
The work required to accelerate a 30.0 kg desk from 0.50 m/s to 1.00 m/s is 11.3 Joules, derived from the change in kinetic energy.
Step-by-step explanation:
The student has inquired how much work must be done to accelerate a 30.0 kg desk from 0.50 m/s to 1.00 m/s. To find the answer, we need to use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy (KE). The kinetic energy of an object is given by KE = 1/2 * m * v^2, where m is the mass and v is the velocity.
The initial kinetic energy, when the desk is moving at 0.50 m/s, is KE_initial = 1/2 * 30.0 kg * (0.50 m/s)^2. The final kinetic energy, when the desk's speed is 1.00 m/s, will be KE_final = 1/2 * 30.0 kg * (1.00 m/s)^2. Therefore, the work done to accelerate the desk is equal to the change in kinetic energy, which is KE_final - KE_initial.
Calculating the value, we find:
KE_initial = 1/2 * 30.0 kg * (0.50 m/s)^2 = 3.75 J
KE_final = 1/2 * 30.0 kg * (1.00 m/s)^2 = 15 J
The work done to accelerate the desk is therefore 15 J - 3.75 J = 11.25 J. Rounding to the nearest tenth of a Joule gives us 11.3 J.