This question involves the concepts of work, the law of conservation of energy, frictional force, normal reaction, and kinetic energy.
a) Work done by the force is "8038.46 J".
b) Work done by the frictional force is "517.96 J".
c) Change in kinetic energy is "7250.5 J"
d) Speed of the crate after covering 9.28 m distance is "27.63 m/s".
a)
The work done by the applied force (188 N) is given as follows:
W = Fd Cosθ
where,
W = Work done = ?
F = Applied force = 188 N
d = distance moved = 48.7 m
θ = Angle between applied force and horizontal surface = 28.6°
Therefore,
W = (188 N)(48.7 m) Cos 28.6°
W = 8038.46 J
b)
Writing equilibrium equation in the vertical direction:
[tex]W = N + FSin\theta[/tex]
where,
W = weight of the block = mg = (19.7 kg)(9.81 m/s²) = 193.26 N
N = Normal Reaction Force = ?
F = Applied force = 188 N
Therefore,
[tex]193.26\ N = N+(188\ N)Sin\ 28.6^o[/tex]
N = 103.26 Newtons
Now, frictional force can be given as follows:
[tex]f =\mu N[/tex]
where,
f = frictional force = ?
μ = coefficient of friction = 0.103
Therefore,
[tex]f=(0.103)(103.26\ N)[/tex]
f = 10.64 N
Thus the work done by this frictional force can be calculated as follows:
[tex]W_f=fd\\W_f=(10.64\ N)(48.7\ m)\\W_f=517.96\ J[/tex]
c)
From the law of conservation of energy:
[tex]Change\ in\ Kinetic\ Energy = Net\ Work\\\Delta K.E=W-W_f = 8038.46\ J-517.96\ J\\[/tex]
ΔK.E = 7520.5 J
d)
The final speed of the crate can be given by the following formula:
[tex]\Delta K.E = 7520.5 J\\\\\frac{1}{2}m(v_f^2-v_i^2)=7520.5\ J\\\\[/tex]
where,
m = mass = 19.7 kg
vf = final speed = ?
vi = initial speed = 0 m/s
Therefore,
[tex]\frac{1}{2}(19.7\ kg)(v_f^2-(0\ m/s)^2) = 7520.5\ J\\\\v_f^2=\frac{(2)(7520.5\ J)}{19.7\ kg}\\\\v_f=\sqrt{763.5 m^2/s^2}\\\\v_f = 27.63\ m/s[/tex]
Learn more about the law of conservation of energy here:
brainly.com/question/20971995?referrer=searchResults
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