At the beginning of a new school term, a student moves a box of books by attaching a rope to the box and pulling with a force of 92 N at an angle of 30 as shown in the figure below. The box of books has a mass of 22.0 kg, and the coefficient of kinetic friction between the bottom of the box and floor is 0.5.
a) Find the acceleration of the box.
b) If the box is at rest to start with, what is its speed after it has traveled 2.00 m?
c) How much time does it take to pull the box this distance?
Please show how you got there! This problem has stumped me..
Wb = m*g = 22kg * 9.8N/kg = 215.6 N. =
Wt. of box.
Fb = 215.6N @ 0o. = = Force of box.
Fp = 215.6*sin(0) = 0 = Force parallel to floor.
Fv = 215.6*cos(0) - 92*sin30 = 169.6 =
Force perpendicular to floor=The Normal.
Fk = u*Fv = 0.5 * 169.6=84.8 N. = Force
of kinetic friction.
a. Fap*cos30-Fp-Fk = m*a.
92*cos30-0-84.8 = 22a
22a = -5.12
a = -0.23 m/s^2. This should not be
negative. Make sure all given INFO is
correct.
b. V ^2 = Vo^2 + 2a*d
To solve this problem, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration (F = ma).
a) Find the acceleration of the box:
First, we need to analyze the forces acting on the box. There are two forces at play: the applied force (92 N) and the force of kinetic friction.
The component of the applied force in the horizontal direction can be found using trigonometry:
F_horizontal = F_applied * cos θ
F_horizontal = 92 N * cos 30°
F_horizontal = 92 N * 0.866 (rounded)
F_horizontal = 79.712 N (rounded)
The force of kinetic friction can be calculated using the equation:
F_friction = μ * N
where μ is the coefficient of kinetic friction and N is the normal force.
The normal force can be found as the weight of the box (mg) in the absence of vertical acceleration:
N = mg
N = 22.0 kg * 9.8 m/s²
N = 215.6 N (rounded)
Now we can calculate the force of kinetic friction:
F_friction = 0.5 * 215.6 N
F_friction = 107.8 N (rounded)
Since the applied force is greater than the force of kinetic friction, there is a net force in the horizontal direction:
Net force = F_horizontal - F_friction
Net force = 79.712 N - 107.8 N
Net force = -28.088 N (rounded)
The negative sign indicates that the net force is in the opposite direction to the applied force.
Now we can find the acceleration using Newton's second law:
Net force = ma
-28.088 N = 22.0 kg * a
a = -1.276 m/s² (rounded)
Therefore, the acceleration of the box is approximately -1.276 m/s².
b) If the box is at rest to start with, what is its speed after it has traveled 2.00 m:
To determine the speed of the box, we can use the kinematic equation:
v² = u² + 2as
where v is the final velocity, u is the initial velocity (0 m/s since the box starts at rest), a is the acceleration, and s is the distance traveled.
Rearranging the equation, we have:
v = √(u² + 2as)
v = √(0² + 2 * -1.276 m/s² * 2.00 m)
v = √(0 + (-5.104 m²/s²))
v = √(-5.104 m²/s²) (taking the square root of a negative value gives us an imaginary number)
Since the acceleration is negative, the box will not have any real speed after traveling 2.00 m. This means the box will not move in the given scenario.
c) How much time does it take to pull the box this distance:
If the box does not move, it means it does not experience any displacement and the time taken to pull the box is undefined.