A crate of mass 10.2 kg is pulled up a rough incline with an initial speed of 1.56 m/s. The pulling force is 92 N parallel to the incline, which makes an angle of 19.0° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.04 m.
(a) How much work is done by the gravitational force on the
crate?
J
(b) Determine the increase in internal energy of the crate–incline
system owing to friction.
J
(c) How much work is done by the 92-N force on the crate?
J
(d) What is the change in kinetic energy of the crate?
J
(e) What is the speed of the crate after being pulled 5.04 m?
m/s
acceleration of the object
a= F/m - u g cos x - g sin x
=92/10.2 - 9.8* (0.4* cos 19 + sin 19)
= 9.02 - 6.897
a= 2.122 m/s^2
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a) workdone by gravitational force
W= mg sin x * d= 10.2*9.8* 5.04 sin 19= - 164.02 J
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b) workdone to change its internal energy
W= 0.4*10.2*9.8 cos 19* 5.04= - 190.54 J
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c) workdone by applied force
W= 92*5.04= 463.68 J
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d) using 3rd equation of motion
V^2= 2* 2.122* 5.04
V= 4.625 m/s
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Comment in case any doubt.. good luck
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