Respuesta :
A) 569.2 J
The work done on the suitcase by the tension force is given by
[tex]W=Fd cos \theta[/tex]
where:
F = T = 186.0 N is the magnitude of the tension
d = 3.20 m is the displacement (the length of the ramp)
[tex]\theta = 17.0^{\circ}[/tex] is the angle between the direction of the force and the displacement of the suitcase
Substituting into the formula, we find:
[tex]W=(186.0 N)(3.20 m) cos 17.0^{\circ}=569.2 J[/tex]
B) -374.4 J
In this case, we need to consider the component of the force of gravity acting along the direction of the ramp, which is given by
[tex]W_p = mg sin \theta=(24.60 kg)(9.81 m/s^2)(sin 29.0^{\circ})=117.0 N[/tex]
and the displacement is the length of the ramp, so the work done by the force of gravity is
[tex]W=W_p d cos \theta=(117.0 N)(3.20 m)(cos 180^{\circ})=-374.4 J[/tex]
And the work is negative because the direction of the component of the force of gravity parallel to the ramp (down the ramp) is opposite to the displacement (up the ramp)
C) 0
In this case, the normal force is perpendicular to the surface of the ramp: this means that the normal force is perpendicular to the displacement of the suitcase, so in the formula
[tex]W=Fd cos \theta[/tex]
[tex]\theta=90^{\circ}, cos 90^{\circ}=0[/tex], so the work done is zero.
D) -182.4 J
The frictional force is given by the product between the coefficient of friction and the normal reaction of the ramp on the suitcase, so it is
[tex]F_f = \mu N = \mu mg cos \theta =(0.270)(24.60 kg)(9.81 m/s^2)(cos 29.0^{\circ})=57.0 N[/tex]
The displacement is the length of the ramp, so the work done by the frictional force is
[tex]W=Fdcos\theta=(57.0 N)(3.20 m)(cos 180^{\circ})=-182.4 J[/tex]
and the angle is [tex]180^{\circ}[/tex] because the frictional force and the displacement have opposite directions, so the work done is negative.
E) 182.4 J
The increase in thermal energy present in the suitcase and in the surface of the ramp is simply equal to the work done by the frictional force (which is converted, in fact, into thermal energy), so it is equal to
E = 182.4 J
F) 12.4 J
The change in kinetic energy of the suitcase will be equal to the work done by the force pulling the suitcase up the ramp, minus the work done against it by gravity and by the frictional force:
[tex]\Delta K=569.2 J-374.4 J-182.4 J=12.4 J[/tex]
The values of the work done by the student, while moving out of his old apartment for difference case, is,
- A.) Work done on the suitcase by the tension force during the pull-up the ramp is 569.2 J.
- B.) Work done by the force of gravity on the suitcase by the Earth during the pull-up the ramp is -374.4 J.
- C.) Work done by the normal force on the suitcase by the surface of the incline during the pull-up the ramp is 0.
- D.) Work done by the frictional force on the suitcase by the ramp during the pull-up of the ramp is -182.4 J.
- E.) Increase in the thermal energy present in the suitcase and in the surface of the ramp is 182.4 J.
- F) The change in the kinetic energy of the suitcase as a result of the pull-up the incline is 12.4 J.
What is work done?
Work done is the force applied on a body to move it over a distance. Work done for inclined plane can be given as,
[tex]W=Fd\cos\theta[/tex]
Here (F) is the magnitude of force and (d) is the distance traveled.
- A.) Work done on the suitcase by the tension force during the pull-up the ramp-
The force (tension) is 186 N and the length of the ramp is 3.2 meters. As, the angle between the strap and the incline is 17.0. Thus, the work done is,
[tex]W=(186)(3.2)\cos (17)\\W=569.2\rm J[/tex]
- B.) Work done by the force of gravity on the suitcase by the Earth during the pull-up the ramp-
The force of gravity acting along the direction of the ramp is given by,
[tex]F=24.60\times9.81\times \sin (29)\\F=117\rm N[/tex]
Thus, the work done in the opposite direction is,
[tex]W=(117)(3.2)\cos (180)\\W=-374.4\rm J[/tex]
- C.) Work done by the normal force on the suitcase by the surface of the incline during the pull-up the ramp-
For this situation, the perpendicular force acts with angle equal to 90 degree and the value of cos 90 is zero.Thus, the work done should is also zero.
- D.) Work done by the frictional force on the suitcase by the ramp during the pull-up the ramp-
The work done of frictional force can be given as,
[tex]W=\mu (mgcos\theta)d\cos \theta[/tex]
Put the values as,
[tex]W=0.270 (24.60\times9.81\times \cos (29))(3.2)\cos (180)\\W=-182.4 \rm J[/tex]
- E.) Increase in the thermal energy present in the suitcase and in the surface of the ramp-
The Increase in the thermal energy present in the suitcase produced by the frictional force. Thus, the increase in the thermal energy is 182.4 J
- F) The change in the kinetic energy of the suitcase as a result of the pull-up the incline-
The change in kinetic energy is the difference of pulling force of the suitcase and the sum of gravitational force and friction force, which opposes the pulling force. Therefore,
[tex]\Delta KE=569.2-(374.4+182.4)\\\Delta KE=12.4\rm J[/tex]
Thus, the values of the work done for the difference cases is obtained above.
Learn more about the work done here;
https://brainly.com/question/25573309
