A luggage handler pulls a suitcase of mass 18.5 kg up a ramp inclined at an angl
ID: 1704327 • Letter: A
Question
A luggage handler pulls a suitcase of mass 18.5 kg up a ramp inclined at an angle 24.0^circ above the horizontal by a F of magnitude 149 N that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is 0.262. The suitcase travels a distance 4.10 m along the ramp
a) Calculate the work done on the suitcase by the force .
b) Calculate the work done on the suitcase by the gravitational force.
c) Calculate the work done on the suitcase by the normal force.
d) Calculate the work done on the suitcase by the friction force.
e) Calculate the total work done on the suitcase.
f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 4.10 m along the ramp?
Explanation / Answer
Given:
mass of the suitcase = 18.5 kg
angle = 24^0
horizontal force F = 149 N
coefficient k = 0.262
(a)
work done by force W = mgsin*s
= 18.5*9.8*sin24*4.1
= 302 J
(b)
work done by gravitational force = -mgcos*s
= -18.5*9.8*cos24*4.1
= 679 J
(c)
work done by Normal force N =0
(d)
work done by a frictional force fk= kmgcos*s
= 0.262*18.5*9.8*cos24*4.1
= 135.8 J
(e)
Net work done = 302-679+135.8 J
= -512.8
(f)
acceleration a = g(sin+kcos)
= 9.8*(sin24+0.262 cos24)
= 6.33 m/s^2
v = sqrt(2as)
= sqrt(2*6.33*4.1)
= 7.2 m/s
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.