Respuesta :
A) 635.7 J
B) -427.4 J
C) 0 J
D) -190.1 J
E) +18.2 J
F) 1.47 m/s
Explanation:
A)
The work done by a force on an object is given by
[tex]W=F_p d[/tex]
where
[tex]F_p[/tex] is the component of the force in the direction of the displacement
d is the displacement of the object
In this problem:
F = 163 N is the magnitude of the force, which acts along the ramp; so, the force is parallel to the displacement,
[tex]F_p = F = 163 N[/tex]
d = 3.90 m is the displacement of the suitcase
Therefore, the work done by force F on the suitcase is
[tex]W_F=(163)(3.90)=635.7 J[/tex]
B)
In order to find the work done by the gravitational force, we use again the equation
[tex]W=F_p d[/tex]
In this case, we need to us the component of the gravitational force parallel to the ramp; this is given by
[tex]F_p = mg sin \theta[/tex]
where
m = 20.0 kg is the mass of the suitcase
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
[tex]\theta=34.0^{\circ}[/tex] is the angle of the ramp
Moreover, this force acts down along the ramp, while the displacement is up along the ramp, so the work done will have a negative sign; so, the work done by the gravitational force is:
[tex]W_g = -mg sin \theta d=-(20.0)(9.8)(sin 34^{\circ})(3.90)=-427.4 J[/tex]
C)
In order to find the work done by the normal force, we use again the equation
[tex]W=F_p d[/tex]
Where [tex]F_p[/tex] is the component of the normal force parallel to the ramp.
The normal force is the reaction force exerted by the surface of the ramp on the suitcase: it acts perpendicular to the ramp.
This means that the component of the normal force in the direction parallel to the ramp is zero:
[tex]F_p = 0[/tex]
And therefore, the work done by the normal force on the suitcase is simply zero:
[tex]W_N=0[/tex]
D)
Here the frictional force acts along the ramp, but in the direction opposite to the motion of the suitcase. Therefore, the work done by the frictional force will have a negative sign:
[tex]W_f=-F_f d[/tex] (1)
Where the frictional force can be written as
[tex]F_f=\mu_k N[/tex]
where
[tex]\mu_K=0.300[/tex] is the coefficient of kinetic friction
N is the normal force
The normal force is equal to the component of the weight perpendicular to the ramp, so:
[tex]N=mg cos \theta[/tex]
Substituting everything into (1), we find:
[tex]W_f=-\mu_k mg cos \theta d[/tex]
And therefore, the work done by friction is:
[tex]W_f=-(0.300)(20.0)(9.8)(cos 34^{\circ})(3.90)=-190.1 J[/tex]
E)
The total work done on the suitcase is the algebraic sum of the works done by each force on the suitcase.
From part A-D, we have:
[tex]W_F=+635.7 J[/tex] is the work done by the force F
[tex]W_g=-427.4 J[/tex] is the work done by the gravitational force
[tex]W_N=0 J[/tex] is the work done by the normal force
[tex]W_f=-190.1 J[/tex] is the work done by the friction force
Therefore, the total work done on the suitcase is:
[tex]W=W_F+W_g+W_N+W_f=+635.7+(-427.4)+0+(-190.1)=+18.2 J[/tex]
F)
The total work done by the forces on the suitcase along 3.90 m was
[tex]W=+18.2 J[/tex]
Since the work done by each force is proportional to the distance covered by the suitcase, the total work done along a distance of 4.60 m will be
[tex]W'=(+18.2) \cdot \frac{4.60}{3.90}=+21.5 J[/tex]
According to the work-energy theorem, the work done on the suitcase is equal to its change in kinetic energy; since the suitcase starts from rest, its initial kinetic energy is zero, so the work done is simply equal to the final kinetic energy:
[tex]W'=\frac{1}{2}mv^2[/tex]
where
m = 20.0 kg is the mass
v is the final velocity of the suitcase
And solvign for v, we find
[tex]v=\sqrt{\frac{2W'}{m}}=\sqrt{\frac{2(21.5)}{20.0}}=1.47 m/s[/tex]
