Respuesta :
(a) 73.5 N
The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law
[tex]\sum F = ma[/tex]
the resultant of the forces must be zero: [tex]\sum F = 0[/tex] (1)
The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:
[tex]F[/tex], the push applied by the worker
[tex]F_f=-\mu mg[/tex], the force of friction, with [tex]\mu=0.25[/tex] being the coefficient of friction, [tex]m=30.0 kg[/tex] being the mass of the crate, and [tex]g=9.8 m/s^2[/tex]. The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes
[tex]F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N[/tex]
So, this is the force that the worker must apply.
(b) 330.8 J
The work done by the pushing force of the worker on the crate is given by:
[tex]W=Fd cos \theta[/tex]
where
F = 73.5 N is the force
d = 4.5 m is the displacement
[tex]\theta=0^{\circ}[/tex] is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)
Substituting, we have
[tex]W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J[/tex]
(c) -330.8 J
To calculate the work done by friction, we apply the same formula:
[tex]W=F_f d cos \theta[/tex]
where
[tex]F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N[/tex] is the magnitude of the force of friction
d = 4.5 m is the displacement
[tex]\theta=180^{\circ}[/tex] is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)
Substituting, we find
[tex]W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J[/tex]
So, the work done by friction is negative.
(d) 0 J
As before, the work done by any force on the crate is
[tex]W=F_f d cos \theta[/tex]
We notice that both gravity and normal force are perpendicular to the displacement: therefore, [tex]\theta=90^{circ}[/tex], and so
[tex]cos \theta=0[/tex]
which means that the work done by both forces is zero.
(e) 0 J
The total work done on the crate is the sum of the work done by the four forces acting on it, so:
[tex]W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0[/tex]
And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.
All the forces applied by the worker to push the crate at a distance of 4.5 m along the level floor,
- (a) Magnitude of force must the worker apply is 330.8 J.
- (b) Work done on the crate by this force is -330.8 J.
- (c) Work done on the crate by friction is 0.
- (d) Work done on the crate by the normal force, By gravity is 0.
- (e) The total work done on the crate is 0.
What is work done?
Work done is the force applied on a body to move it over a distance. Work done to move a body can be given as,
[tex]W=Fd\cos \theta[/tex]
Here, (F) is the magnitude of force and (d) is the distance traveled.
- (a) Magnitude of force must the worker apply-
The force needed to apply on the crate, by the worker to move it is equal to the friction force which is opposing the crate to move. Thus, the force required is,
[tex]F=\mu mg\\F=0.25\times30\times9.8\\F=73.5\rm N[/tex]
- (b) Work done on the crate by this force-
Now the work done can be found out by multiplying the force to the distance object move as,
[tex]W=73.5\times4.5\\W=330.8\rm J[/tex]
- (c) Work done on the crate by friction-
Now the work done on the crate by friction is equal and opposite to the work done on the crate by the force worker required to move the crate. Therefore,
[tex]W_f=-W\\W_f=-330.8\rm J[/tex]
- (d) Work done on the crate by the normal force, By gravity-
The force acting on the body is perpendicular to the gravitational force. Therefore, the component of the force in the direction of 90 degrees. Thus, the work done will be,
[tex]W=Fd\cos(90)\\W=0[/tex]
- (e) The total work done on the crate-
Now the total work done on the crate is the sum of all the force as,
[tex]W_T=330.8+(-330.8)+0+0\\W_T=0[/tex]
Thus, all the forces applied by the worker to push the crate at a distance of 4.5 m along the level floor,
- (a) Magnitude of force must the worker apply is 330.8 J.
- (b) Work done on the crate by this force is -330.8 J.
- (c) Work done on the crate by friction is 0.
- (d) Work done on the crate by the normal force, By gravity is 0.
- (e) The total work done on the crate is 0.
Learn more about the work done here;
https://brainly.com/question/25573309
