Answer:
(a) work done by the gravitational force on the crate is -167.59J
(b) the increase in internal energy of the crate-incline system owing to friction is 184.2 J
(c) work done by the 100- N force on the crate is 469.85 J
(d) the change in kinetic energy of the crate is 118.06 J
(e) the speed of the crate after being pulled 5.0 m is 5.09 m/s
Explanation:
Given;
mass of the crate, m = 10.0 kg
initial speed of the crate, u = 1.50 m/s
parallel force on the crate, F = 100N
angle of inclination of the force, θ = 20°
coefficient of friction, μ = 0.4
distance moved by the crate, h = 5.0m
Part (a) work done by the gravitational force on the crate:
This work is due to vertical displacement of the crate = h*sinθ,
W = -(mgh*sinθ)= -(10*9.8*5.0*sin(20))= -167.59 J
This work is negative because it is acting against the crate.
Part (b) the increase in internal energy of the crate-incline system owing to friction.
Magnitude of frictional force = μ*N
where;
N is the normal reaction = mgcosθ
Magnitude of frictional force = μ*mgcosθ
= 0.4*10*9.8*cos(20) = 36.84 N
Thus, increase in internal energy of the the crate-incline system owing to friction = Magnitude of frictional force X h
= 36.84N x 5m
= 184.2 J
Part (c) work done by the 100- N force on the crate
W = Fd*cosθ = 100*5*cos(20) = 469.85 J
Part (d) the change in kinetic energy of the crate:
ΔK.E = W.force - W.gravity - W.friction
ΔK.E = 469.85 J - 167.59 - 184.2 J = 118.06 J
Part (e) the speed of the crate after being pulled 5.0 m
Δ.KE [tex]= K.E_f-K.E_i\\\\[/tex]
[tex]\delta K.E= \frac{1}{2}mv_f^2 -\frac{1}{2}mv_i^2\\\\\frac{1}{2}mv_f^2 = \delta K.E + \frac{1}{2}mv_i^2\\\\v_f = \sqrt{\frac{2\delta K.E}{m} +v_i^2}\\\\v_f = \sqrt{\frac{2*118.06}{10} +(1.5)^2} =5.09 m/s[/tex]
v = 5.09 m/s
This question involves the concepts of the law of conservation of energy, potential energy, kinetic energy, and frictional force.
(a) Work done by the gravitational force on the crate is "167.8 J".
(b) The increase in the internal energy of the crate-incline system owing to the frictional force is "184.4 J".
(c) "500 J" work is done by the 100 N force.
(d) The change in kinetic energy of the crate is "147.8 J".
(e) The speed of the crate after being pulled 5 m is "5.64 m/s".
(a)
The work done by gravitational force must be equal to the increase in gravitational potential energy of the crate:
[tex]W_g=mgdSin\theta[/tex]
where,
[tex]W_g[/tex] = Work done by the gravitational force = ?
m = mass of crate = 10 kg
g = acceleration due to gravity = 9.81 m/s²
d = distance moved parallel to incline plane = 5 mm = 5 m
θ = angle of inclinition = 20°
Therefore,
[tex]W_g = (10\ kg)(9.81\ m/s^2)(5\ m)Sin20^o\\W_g = 167.8\ J[/tex]
(b)
The increase in internal energy owing to frictional force must be equal to the work done by the frictional force:
[tex]\Delta U = (Frictional Force)(d)\\\Delta U = (\mu N)(d) \\\Delta U = (\mu mgCos\theta)(d)[/tex]
where,
μ = coefficient of friction = 0.4
Therefore,
[tex]\Delta U=(0.4)(10\ kg)(9.81\ m/s^2)(Cos20^o)(0.005\ m)[/tex]
ΔU = 184.4 J
(c)
Work done by the 100 N force is simply given as follows:
[tex]W = Fd Cos\theta[/tex]
where,
θ = the angle between motion and force = 0° (since the force is parallel to inclined plane)
Therefore,
W = (100 N)(5 m)Cos 0°
W = 500 J
(d)
The change in kinetic energy of the crate can be given by applying the law of conservation of energy to this situation:
[tex]Change\ in\ Kinetic\ Energy = \Delta K.E=W-W_g-\Delta U\\\Delta K.E=500\ J - 167.8\ J-184.4\ J\\[/tex]
ΔK.E = 147.8 J
(e)
The change in kinetic energy is given as follows:
[tex]\Delta K.E =\frac{1}{2}m(v_f^2-v_i^2)\\\\147.8\ J = \frac{1}{2}(10\ kg)((v_f)^2-(1.5\ m/s)^2)\\\\(5\ kg)v_f^2=147.8\ J+11.25\ J\\\\v_f=\sqrt{\frac{159.05\ J}{5\ kg}}\\\\v_f=5.64\ m/s[/tex]
Learn more about the law of conservation of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture explains the law of conservation of energy.
