Question

Your friend provides a solution to the following problem. Evaluate his solution. The problem: A 400-kg motorcycle, including the driver, travels up a 10.0-m-long ramp inclined 30.0∘ above the paved horizontal surface holding the ramp. The cycle leaves the ramp at speed 20.0 m/s. Determine the cycle's speed just before it lands on the paved surface. Your friend's solution: (1/2)(400kg)(20m/s)=(400kg)(9.8m/s2)(10m)+(1/2)(400kg)v2 v=−13.2m/s

Answer 1

Answer:

v=22m/s

Explanation:

In the friend's solution there are 2 mistakes:

First velocity is not squared on LHS as it is Kinetic energy

Secondly work which is gravitational Potential energy is not calculated accurately because the inclined angle is not considered.

Angle between gravity vector and displacement vector is 90+30=120 degree

Correct Solution:

Work=-400kg(9.8m/s2)(10cos120)

$\frac{1}{2} (400kg)*(20^{2})= -400kg(9.8m/s2)(10cos120)+\frac{1}{2} (400kg)*(v^{2})$

$80000=-19600+200*v^{2}$

v=22m/s