suppose that a 70.0kg box is pulled by a 400.N force at an angle of 30 degrees to the horizontal. the coefficient of sliding friction is 0.500. what is the acceleration of the box?
To find the acceleration of the box, we need to calculate the net force acting on it. Here are the steps to solve this problem:
Step 1: Resolve the 400 N force into its vertical and horizontal components.
The vertical component can be found by multiplying the force by the sine of the angle:
Vertical Component = 400 N * sin(30°) = 200 N.
The horizontal component can be found by multiplying the force by the cosine of the angle:
Horizontal Component = 400 N * cos(30°) = 346.4 N.
Step 2: Calculate the force of friction.
The force of friction can be found by multiplying the coefficient of sliding friction (μ) by the normal force (which equals the weight of the box):
Weight of the box = mass * gravity
Weight of the box = 70.0 kg * 9.8 m/s^2 = 686 N.
Force of friction = μ * normal force = 0.500 * 686 N = 343 N.
Step 3: Calculate the net force.
The net force is the difference between the horizontal component of the applied force and the force of friction:
Net Force = Horizontal Component - Force of Friction
Net Force = 346.4 N - 343 N = 3.4 N.
Step 4: Calculate the acceleration.
The acceleration can be found using Newton's second law: F = m * a, where F is the net force and m is the mass of the box:
Net Force = m * a
3.4 N = 70.0 kg * a.
Simplify and solve for a:
a = 3.4 N / 70.0 kg = 0.0486 m/s^2.
Therefore, the acceleration of the box is approximately 0.0486 m/s^2.
To find the acceleration of the box, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a).
First, let's resolve the force into horizontal and vertical components. The horizontal component is given by Fh = F * cos(theta) and the vertical component is Fv = F * sin(theta), where F is the applied force and theta is the angle.
Given:
Mass of the box (m) = 70.0 kg
Applied force (F) = 400 N
Angle (theta) = 30 degrees
Coefficient of sliding friction (μ) = 0.500
Horizontal component (Fh) = F * cos(theta)
Fh = 400 N * cos(30°)
Fh = 400 N * √3/2
Fh ≈ 400 N * 0.866
Fh ≈ 346.41 N (rounded to two decimal places)
Now, let's calculate the frictional force. The frictional force (Ff) can be determined using the equation Ff = μ * Fn, where Fn is the normal force. The normal force (N) is the force exerted by the surface and is equal to the weight of the object.
Weight of the box (W) = m * g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Weight of the box (W) = 70.0 kg * 9.8 m/s²
W ≈ 686 N (rounded to two decimal places)
Normal force (Fn) = W
Fn = 686 N
Frictional force (Ff) = μ * Fn
Ff = 0.500 * 686 N
Ff = 343 N
Now, let's calculate the net force acting on the box in the horizontal direction.
Net force (Fnet) = Fh - Ff
Fnet = 346.41 N - 343 N
Fnet ≈ 3.41 N (rounded to two decimal places)
Finally, we can determine the acceleration (a) using the equation F = m * a.
Fnet = m * a
3.41 N = 70.0 kg * a
Now, solve for acceleration:
a = 3.41 N / 70.0 kg
a ≈ 0.0490 m/s² (rounded to four decimal places)
Therefore, the acceleration of the box is approximately 0.0490 m/s².
the horizontal force is 400cos30
vertical force=400Sin30
net apparent weight=70g-400sin30
acceleration=nethorizonalforce/mass
= (400cos30-mu(70g-400sin30)/70