Respuesta :
a) 73.5 N
b) 551.3 J
c) -551.3 J
d) 0 J
e) 0 J
f) 0 J
g) 0 J
Explanation:
a)
There are two forces acting on the crate:
- The push of the worker, F, in the forward direction
- The frictional force, [tex]F_f=\mu mg[/tex], in the backward direction, where
[tex]\mu=0.15[/tex] is the coefficient of friction
m = 50 kg is the mass of the crate
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
According to Newton's second law of motion, the net force on the crate must be equal to the product of mass and acceleration, so:
[tex]F-F_f=ma[/tex]
However, the crate here is moving with constant velocity, so its acceleration is zero:
[tex]a=0[/tex]
So the previous equation becomes:
[tex]F-F_f=0[/tex]
And we can find the magnitude of the applied force:
[tex]F=F_f=\mu mg=(0.15)(50)(9.8)=73.5 N[/tex]
b)
The work done by the applied force on the crate is
[tex]W_F=Fd cos \theta[/tex]
where:
F is the magnitude of the force
d is the displacement of the crate
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
Here we have:
F = 73.5 N
d = 7.5 m
[tex]\theta=0^{\circ}[/tex] (the force is applied in the same direction as the displacement)
Therefore,
[tex]W_F=(73.5)(7.5)(cos 0^{\circ})=551.3 J[/tex]
c)
The work done by friction on the crate is:
[tex]W_{F_f}=F_f d cos \theta[/tex]
where in this case:
[tex]F_f=73.5 N[/tex] is the magnitude of the force of friction
d = 7.5 m is the displacement of the crate
[tex]\theta=180^{\circ}[/tex], because the displacement is forward and the force of friction is backward, so they are in opposite direction
Therefore, the work done by the force of friction is:
[tex]W_{F_f}=(73.5)(7.5)(cos 180^{\circ})=-551.3 J[/tex]
d)
To find the normal force, we analyze the situation of the force along the vertical direction.
We have two forces on the vertical direction:
- The normal force, N, upward
- The force of gravity, [tex]mg[/tex], downward, where
m = 50 kg is the mass of the crate
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
Since the crate is in equilibrium in this direction, the vertical acceleration is zero, so the two forces balance each other:
[tex]N-mg=0\\N=mg=(50)(9.8)=490 N[/tex]
The work done by the normal force is:
[tex]W_N=Nd cos \theta[/tex]
In this case, [tex]\theta=90^{\circ}[/tex], since the normal force is perpendicular to the displacement of the crate; therefore, the work done is
[tex]W_N=(490)(7.5)(cos 90^{\circ})=0[/tex]
e)
The work done by the gravitational force is:
[tex]W_g=F_g d cos \theta[/tex]
where:
[tex]F_g=mg=(50)(9.8)=490 N[/tex] is the gravitational force
d = 7.5 m is the displacement of the crate
[tex]\theta=90^{\circ}[/tex] is the angle between the direction of the gravitational force (downward) and the displacement (forward)
Therefore, the work done by gravity is
[tex]W_g=(490)(7.5)(cos 90^{\circ})=0 J[/tex]
f)
The total work done on the crate can be calculated by adding the work done by each force:
[tex]W=W_F+W_{F_f}+W_N+W_g[/tex]
where we have:
[tex]W_F=+551.3 J[/tex] is the work done by the applied force
[tex]W_{F_f}=-551.3 J[/tex] is the work done by the frictional force
[tex]W_N=0[/tex] is the work done by the normal force
[tex]W_g=0[/tex] is the work done by the force of gravity
Substituting,
[tex]W=+551.3+(-551.3)+0+0=0 J[/tex]
So, the total work is 0 J.
g)
According to the work-energy theorem, the change in kinetic energy of the crate is equal to the work done on it, therefore:
[tex]W=\Delta E_K[/tex]
where
W is the work done on the crate
[tex]\Delta E_K[/tex] is the change in kinetic energy of the crate
In this problem, we have:
[tex]W=0[/tex] (total work done on the crate is zero)
Therefore, the change in kinetic energy of the crate is:
[tex]\Delta E_K = W = 0[/tex]
