The net increase in potential energy by the crate is negative 49.4 Joules. That is, ΔP.E = - 49.4 J
Given that a 40N box is pushed 5 m up a ramp. The angle of the ramp is 37° above the horizontal.
We can find the height h of the ramp with the equation below;
h = 5sin 37
h = 3.009 m
Mass of the box = 40/9.8 = 4.082 kg
The maximum energy attained = mgh
Maximum Energy = 4.082 x 9.8 x 3.009
E = 120.36 J
The maximum K.E = Maximum P.E
1/2m[tex]U^{2}[/tex] = 120.36
m[tex]U^{2}[/tex] = 240.72
[tex]U^{2}[/tex] = 240.72 / 4.082
[tex]U^{2}[/tex] = 58.97
U = [tex]\sqrt{58.97}[/tex]
U = 7.7 m/s
We can find the acceleration by using third equation of motion
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] - 2aS
Since the box is going to rest, V = 0, and acceleration will be negative
0 = 7.7^2 - 2 x a x 5
10a = 59.29
a = 59.29/10
a = 5.93 m/[tex]s^{2}[/tex]s^2
The applied force F = ma
F = 4.082 x 5.93
F = 24.2 N
Given also that the applied force is parallel to the ramp and opposed by a frictional force of 10.0 N. The net increase in potential energy by the crate will be
ΔP.E = Net work done
ΔP.E = Net force x distance
ΔP.E = (24.2 - 10 - 40sin37) x 5
ΔP.E = - 9.87 x 5
ΔP.E = - 49.4 J
Therefore, The net increase in potential energy by the crate is negative 49.4 Joules.
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