(a) 756.9 J
The work done by a force when moving an object is given by:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement of the object
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this situation,
F = 165 N
d = 5.60 m
[tex]\theta=35.0^{\circ}[/tex]
so the work done by the tension is
[tex]W=(165 N)(5.60 m)cos 35.0^{\circ}=756.9 J[/tex]
(b) 0.646
The crate is moving at constant speed: this means that the acceleration of the crate is zero, so the net force on the crate is also zero.
There are only two forces acting on the crate along the horizontal direction:
- The horizontal component of the tension, [tex]Tcos \theta[/tex], forward
- The frictional force, [tex]-\mu N[/tex], backward, with [tex]\mu[/tex] being the coefficient of kinetic friction, N being the normal reaction of the surface on the crate
The normal reaction is the sum of the weight of the crate + the vertical component of the tension, so
[tex]N=mg-T sin \theta = (31.0 kg)(9.8 m/s^2) - (165 N)sin 35.0^{\circ}=209.2 N[/tex]
Since the net force is zero, we have
[tex]T cos \theta-\mu N =0[/tex]
where
T = 165.0 N
[tex]\theta=35.0^{\circ}[/tex]
N = 209.2 N
Solving for [tex]\mu[/tex], we find the coefficient of friction:
[tex]\mu = \frac{T cos \theta}{N}=\frac{(165.0 N)(cos 35^{\circ})}{209.2 N}=0.646[/tex]
