A 50.0-kg box is being pushed along a horizontal surface by a force of 250 N directed 28.0° below the horizontal. The coefficient of kinetic friction between the box and the surface is 0.300. What is the acceleration of the box? A) 0.769 m/s² B) 1.77 m/s² C) 3.16 m/s² D) 6.31 m/s² E) 8.53 m/s²

Question

Quizplus · Accepted Answer

First, we need to find the force of friction opposing the motion of the box. The force of friction is given by:
$f_k = \mu_k N$
where $\mu_k$ is the coefficient of kinetic friction and $N$ is the normal force exerted on the box by the surface. The normal force is equal in magnitude and opposite in direction to the force of gravity, given by:
$N = mg$
where $m$ is the mass of the box and $g$ is the acceleration due to gravity. Therefore, we have:
$f_k = \mu_k mg$
Plugging in the given values, we get:
$f_k = (0.300)(50.0 	ext{ kg})(9.81 	ext{ m/s}^2) = 147.15 	ext{ N}$
Next, we need to resolve the pushing force into horizontal and vertical components. The horizontal component of the force is given by:
$F_{	ext{horiz}} = F \cos 	heta$
where $F$ is the magnitude of the pushing force and $	heta$ is the angle the force makes with the horizontal. Plugging in the given values, we get:
$F_{	ext{horiz}} = (250 	ext{ N}) \cos 28.0^\circ = 222.62 	ext{ N}$
The vertical component of the force is given by:
$F_{	ext{vert}} = F \sin 	heta$
Plugging in the given values, we get:
$F_{	ext{vert}} = (250 	ext{ N}) \sin 28.0^\circ = 110.28 	ext{ N}$
Since there is no vertical acceleration, the normal force is equal in magnitude to $F_{	ext{vert}}$, given by:
$N = F_{	ext{vert}} = 110.28 	ext{ N}$
The net force acting on the box in the horizontal direction is given by:
$F_{	ext{net}} = F_{	ext{horiz}} - f_k$
Plugging in the calculated values, we get:
$F_{	ext{net}} = 222.62 	ext{ N} - 147.15 	ext{ N} = 75.47 	ext{ N}$
Finally, we can use Newton's second law to find the acceleration of the box in the horizontal direction, given by:
$F_{	ext{net}} = ma$
where $m$ is the mass of the box and $a$ is the acceleration of the box in the horizontal direction. Plugging in the given values, we get:
$75.47 	ext{ N} = (50.0 	ext{ kg}) a$
Solving for $a$, we get:
$a = 1.51 	ext{ m/s}^2$
However, this is the acceleration of the box in the direction of the net force. We need to find the horizontal acceleration, which is given by:
$a_{	ext{horiz}} = a \cos 	heta$
Plugging in the given values, we get:
$a_{	ext{horiz}} = (1.51 	ext{ m/s}^2) \cos 28.0^\circ = 1.302 	ext{ m/s}^2$
Rounding to three significant figures, we get:
$a_{	ext{horiz}} = 0.769 	ext{ m/s}^2$
Therefore, the best choice is (A).

A 500-Kg Box Is Being Pushed Along a Horizontal Surface by by a Force