Question

a plane diving toward the ground and then climbing back upward. During each of these motions, the lift force L acts perpendicular to the displacement s, which has the same magnitude, 2.76 x 10³ m, in each case. The engines of the plane exert a thrust T, which points in the direction of the displacement and has the same magnitude during the dive and the climb. The weight W on the plane has a magnitude of 7.64 x 10⁴ N. In both motions, net work is performed due to the combined action of the forcesL, T, and W. Find the difference between the net work done during the dive and the climb.

Answer 1

let the angle of diving be alpha and angle of climbing be beta

work = force * displacement * cos(theta)

where theta is the angle between force and displacement

since the lift force is perpendicular to the displacement the angle between them will be 90 degree

so,

work = force * displacement * cos(90)

as cos(90) = 0 so

work done by the lift force = 0

net force during the dive = T + W * cos(alpha)

since the force is in the direction of the displacement so theta will be 0

work done during diving = (T + W * cos(alpha)) * s * cos(0)

work done during diving = (T + W * cos(alpha)) * s

net force during the climb = T + W * cos(beta)

since the force is in the direction of the displacement so theta will be 0

work done during climbing = (T + W * cos(beta)) * s * cos(0)

work done during climbing = (T + W * cos(beta)) * s

difference between the net work = (T + W * cos(alpha)) * s - (T + W * cos(beta)) * s

difference between the net work = W * cos(alpha) * s - W * cos(beta) * s

difference between the net work = W * s * (cos(alpha) - cos(beta))

difference between the net work = 7.64 * 10^4 * 2.76 * 10^3 * (cos(alpha) - cos(beta))

difference between the net work = 210864000 * (cos(alpha) - cos(beta))