A force of 15 N acts on a body with a mass of 7.5 kg for 5 s. Determine the work of this force

A force of 15 N acts on a body with a mass of 7.5 kg for 5 s. Determine the work of this force if the direction of the vectors of displacement and force: a) coincide b) make up an angle of 30`. Calculate the power developed by the force

F = 15 N.

m = 7.5 kg.

t = 5 s.

∠α1 = 00

∠α2 = 300.

A1 -?

N1 -?

A2 -?

N2 -?

The work of force A is determined by the formula: A = F * S * cosα, where F is the force that acts on the body, S is the displacement of the body under the action of the force, ∠α is the angle between force F and displacement S.

Power N is a physical quantity that shows the speed of work: N = A / t.

a = F / m.

S = a * t ^ 2/2 = F * t ^ 2/2 * m.

A = F ^ 2 * cosα * t ^ 2/2 * m.

A1 = F ^ 2 * cosα1 * t ^ 2/2 * m.

A1 = (15 N) ^ 2 * cos00 * (5 s) ^ 2/2 * 7.5 kg = 375 J.

N1 = A1 / t.

N1 = 375 J / 5 s = 75 W.

A2 = F2 * cosα2 * t ^ 2/2 * m.

A2 = (15 N) 2 * cos300 * (5 s) ^ 2/2 * 7.5 kg = 325 J.

N2 = A2 / t.

N2 = 325 J / 5 s = 65 W.

Answer: A1 = 375 J, N1 = 75 W, A2 = 325 J, N2 = 65 W.