1) The work done by tension is 12.0 J
2) The final speed of the box is 1.57 m/s
Explanation:
The work done by a force on an object is given by:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
For the box in this problem, we have:
F = 6.2 N is the force applied (the tension in the string)
d = 193.0 cm = 1.93 m is the displacement
[tex]\theta=0[/tex] assuming that the string is horizontal
Substituting, we find the work done by the tension:
[tex]W=(6.2)(1.93)(cos 0)=12.0 J[/tex]
2)
In order to determine the final speed of the box, we need to determine its acceleration first.
Beside the tension, acting forward, the other force acting horizontally on the box is the force of friction, whose magnitude is
[tex]F_f = \mu mg[/tex]
where
[tex]\mu=0.16[/tex] is the coefficient of friction
m = 2.8 kg is the mass of the box
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Substituting,
[tex]F_f = (0.16)(2.8)(9.8)=4.4 N[/tex]
Therefore, the net force on the box is
[tex]F=6.2 N - 4.4 N = 1.8 N[/tex]
And the acceleration can be found by using Newton's second law:
[tex]F=ma[/tex]
where
F = 1.8 N is the net force
m = 2.8 kg is the mass of the box
a is the acceleration
Solving for a,
[tex]a=\frac{F}{m}=\frac{1.8}{2.8}=0.64 m/s^2[/tex]
Now we can finally find the final speed using the suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v is the final speed
u = 0 is the initial speed
[tex]a=0.64 m/s^2[/tex] is the acceleration
s = 1.93 m is the displacement
Solving for v,
[tex]v=\sqrt{u^2+2as}=\sqrt{0+2(0.64)(1.93)}=1.57 m/s[/tex]
Learn more about work:
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