A 80.0 N box of is pulled 20.0 m up a 30.0 Degree ramp by a force of 115 N, if the coefficient of kinetic friction between the box and the ramp is .22 what is the box's kinetic energy?

Answer 1

Kinetic energy = (1/2)mv^2

m = 80.0 N
v = (115 N * 20.0 m * cos(30.0°)) / (80.0 N * .22)

Kinetic energy = (1/2) * 80.0 N * (115 N * 20.0 m * cos(30.0°))^2 / (80.0 N * .22)^2

Kinetic energy = (1/2) * (115 N * 20.0 m * cos(30.0°))^2 / (.22)^2

Kinetic energy = (1/2) * (115 N * 20.0 m * 0.866)^2 / (.22)^2

Kinetic energy = (1/2) * (1953.2 N * m)^2 / (.22)^2

Kinetic energy = (1/2) * (3.737 x 10^5 N^2 * m^2) / (.22)^2

Kinetic energy = 8.5 x 10^5 J