Question

An unknown force is applied to a 12 kg mass. The force acts at an angle of 30.0 degrees above the horizontal. Determine the force acting if the force acts for a horizontal displacement of 22 meters and increases the 12 kg mass's speed from 11 m/s to 26 m/s.

Answer 1

The mass undergoes a change in kinetic energy of

∆K = 1/2 (12 kg) [(26 m/s)² - (11 m/s)²] = 3330 J

By the work-energy theorem, the total work W performed on the mass is equal to ∆K. Assuming no friction, only the horizontal component of the applied force performs work on the mass.

Let F be the magnitude of this force, so that its horizontal component has mag. F cos(30.0°) = √3/2 F. It acts over a displacement of 22 m and performs 3330 J of work in the process, so that

3330 J = (√3/2 F) (22 m)   ⇒   F = 1110√3/11 N ≈ 175 N