A crate of mass 9.8 kg is pulled up a rough incline with an initial speed of 1.4
ID: 1349363 • Letter: A
Question
A crate of mass 9.8 kg is pulled up a rough incline with an initial speed of 1.46 m/s. The pulling force is 102 N parallel to the incline, which makes an angle of 20.2° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.90 m.
(a) How much work is done by the gravitational force on the crate?
(b) Determine the increase in internal energy of the crate–incline system owing to friction.
(c) How much work is done by the 102-N force on the crate?
(d) What is the change in kinetic energy of the crate?
(e) What is the speed of the crate after being pulled 4.90 m?
Explanation / Answer
given data
mass m=9.8 kg
coefficient of kinetic friction muk=0.4
pulling force F=102N
the initial velocity vi=1.46 m/s
let force of friction f, final velocity vf
a)
work is done by the gravitational force on the crate is
Wg=m*g*h
=9.8*9.8*4.90*sin 20.2
=162.4 J
b)
frictional work = force*distance
=muk*N*d
=muk*(m*g*cos theta)*d
=0.4*(9.8*9.8*cos20.2)*4.9
=176.66 J
c)
work done W = F*d
=102*4.9
=499.8 J
d)
frictional force f= uk*N
= uk*m*g*cos theta
=0.4*9.8*9.8*cos 20.2
=36 N
F - m*g*sin theta - f = m*a
a=(F - m*g*sin theta -f)/(m)
=(102-9.8*9.8 *sin 20.2-36)/9.8
=3.35 m/s^2
from the kinematic equations
vf^2=vi^2+2*a*d
vf^2=1.46^2+2*3.35*4.9
=34.9616
vf=5.91 m/s
the change in kinetic energy of the crate
=0.5*m*(vf^2-vi^2)
=0.5*9.8*(34.9616-2.1316)
=643.46J
e)
the speed of the crate after being pulled is
vf^2=34.9616
vf=5.91 m/s
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