Question

You drag a suitcase of mass 8.2 kg with a force of f at an angle 41.9 ◦ with respect to the horizontal along a surface with kinetic coefficient of friction 0.33. the acceleration of gravity is 9.8 m/s 2 . if the suitcase is moving with constant velocity 1.58 m/s, what is f? answer in units of n.

Answer 1

Answer:

35.6 N

Explanation:

We can consider only the forces acting along the horizontal direction to solve the problem.

There are two forces acting along the horizontal direction:

- The horizontal component of the pushing force, which is given by

$F_x = F cos \theta$

with $\theta=41.9^{\circ}$

- The frictional force, whose magnitude is

$F_f = \mu mg$

where $\mu=0.33$, m=8.2 kg and g=9.8 m/s^2.

The two forces have opposite directions (because the frictional force is always opposite to the motion), and their resultant must be zero, because the suitcase is moving with constant velocity (which means acceleration equals zero, so according to Newton's second law: F=ma, the net force is zero). So we can write:

$F_x - F_f=0\\F_x = F_f\\F cos \theta = \mu mg\\F=\frac{\mu mg}{cos \theta}=\frac{(0.33)(8.2 kg)(9.8 m/s^2)}{cos(41.9^{\circ})}=35.6 N$