(a) 0.71 m/s^2
We are only interested in the horizontal motion of the crate, so there are two forces acting on the crate along this direction:
- The horizontal component of the force exerted by the worker through the rope, which is given by
[tex]F cos \theta[/tex]
where F = 450 N and [tex]\theta=38^{\circ}[/tex]
- The horizontal force of friction that opposes the motion, given by
[tex]F_f = 125 N[/tex]
The two forces have opposite directions, so we must take into account a negative sign. According to Newton's second law, the resultant of these forces is equal to the product between the mass of the crate, m, and its acceleration, a:
[tex]F cos \theta - F_f = ma[/tex]
where m = 325 kg. Re-arranging the equation, we can find the magnitude of the acceleration:
[tex]a=\frac{F cos \theta - F_f}{m}=\frac{(450 N)(cos 38^{\circ})-125 N}{325 kg}=0.71 m/s^2[/tex]
(b) 6.92 m/s^2
In this case, we need to find the mass of the crate first. We know that its weight is given by
[tex]W=mg[/tex]
where g=9.8 m/s^2 is the acceleration due to gravity. Since we know W=325 N, we find
[tex]m=\frac{W}{g}=\frac{325 N}{9.8 m/s^2}=33.2 kg[/tex]
And now we can use the same equation used in part (a) to find the acceleration of the crate:
[tex]a=\frac{F cos \theta - F_f}{m}=\frac{(450 N)(cos 38^{\circ})-125 N}{33.2 kg}=6.92 m/s^2[/tex]
