Question

A 17926-lb truck enters an emergency exit ramp at a speed of 75.6 ft/s. It travels for 6.4 s before its speed is reduced to 30.3 ft/s. Determine the braking force by the truck if the acceleration is constant. (Use Impulse-Momentum concepts.) Assume theta = 21.4 degrees

Answer 1

Answer:

$F_{braking}=337299 pdl$

Explanation:

Impulse-Momentum relation:

$I=\Delta p\\ F_{total}*t=m(v_{f}-v{o})$

$F_{total}=-F_{braking}+mgsin{\theta}$

We solve the equations in order to find the braking force:

$F_{braking}=m(v_{o}-v{f})/t+mgsin{\theta}=17926(75.6-30.3)/6.4+17926*32.17*sin21.4=337299 pdl$