As shown in the figure below, a box of mass
m = 57.0 kg
(initially at rest) is pushed a distance
d = 79.0 m
across a rough warehouse floor by an applied force of
FA = 202 N
directed at an angle of 30.0° below the horizontal. The coefficient of kinetic friction between the floor and the box is 0.100. Determine the following. (For parts (a) through (d), give your answer to the nearest multiple of 10.)
(a) The work done by the applied force:
14,400 J
To determine the following values, we need to use the given data and apply the equations of motion and the laws of physics. Let's break down each part step by step.
(a) The work done by the applied force in moving the box a distance of 79.0 m.
The work done (W) can be calculated using the formula: W = F * d * cos(theta), where F is the applied force, d is the distance, and theta is the angle between the force and the displacement.
Using the given data:
F = 202 N
d = 79.0 m
theta = 30.0°
We can calculate the work done as:
W = 202 N * 79.0 m * cos(30.0°)
Calculating the above expression gives us the value of work done.
(b) The work done by the frictional force.
The work done by the frictional force is given by the formula: W_f = -f * d, where f is the frictional force and d is the distance.
To find the frictional force, we use the equation: f = u * m * g, where u is the coefficient of kinetic friction, m is the mass, and g is the acceleration due to gravity.
Using the given data:
u = 0.100
m = 57.0 kg
g = 9.8 m/s^2
d = 79.0 m
We can find the frictional force using the above equation. Then calculate the work done using the formula mentioned earlier.
(c) The net work done on the box.
The net work done is given by the sum of the work done by the applied force and the work done by the frictional force. In this case, the net work done can be calculated as the sum of the values obtained in parts (a) and (b).
(d) The final speed of the box.
To find the final speed of the box, we can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. Mathematically, this can be expressed as:
W_net = ΔKE
Since the box starts from rest, the initial kinetic energy is zero. Therefore, the net work done is equal to the change in kinetic energy, which can be written as:
W_net = KE_final - KE_initial
Simplifying the equation, we get:
W_net = (1/2) * m * v_final^2 - 0
Solving for v_final, we can find the final speed of the box.
By following these steps and applying the appropriate equations, we can determine each of the requested values.
To solve this problem, we need to consider the forces acting on the box and use Newton's second law of motion:
1. Determine the weight of the box:
The weight of an object is given by the formula:
Weight = mass × acceleration due to gravity
Weight = m × g
where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s²).
Given that the mass of the box is m = 57.0 kg, we can calculate the weight:
Weight = 57.0 kg × 9.8 m/s² = 558.6 N
2. Determine the normal force acting on the box:
The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, the normal force is equal in magnitude and opposite in direction to the weight of the box.
Therefore, the normal force is 558.6 N, acting vertically upwards.
3. Determine the force of friction:
The force of friction can be calculated using the formula:
Force of friction = coefficient of friction × normal force
Given that the coefficient of kinetic friction is 0.100 and the normal force is 558.6 N, we can calculate the force of friction:
Force of friction = 0.100 × 558.6 N = 55.9 N
4. Resolve the applied force into horizontal and vertical components:
The applied force FA can be resolved into horizontal and vertical components using trigonometry. The horizontal component is given by:
Horizontal component of FA = FA × cos(angle)
Given that the angle below the horizontal is 30.0° and the applied force is FA = 202 N, we can calculate the horizontal component:
Horizontal component of FA = 202 N × cos(30.0°) = 175.13 N
The vertical component is given by:
Vertical component of FA = FA × sin(angle)
Given the same values as above, we can calculate the vertical component:
Vertical component of FA = 202 N × sin(30.0°) = 101 N
5. Calculate the net force in the horizontal direction:
The net force in the horizontal direction is the difference between the applied force (horizontal component) and the force of friction:
Net force in the horizontal direction = horizontal component of FA - Force of friction
Using the given values, we can calculate the net force in the horizontal direction:
Net force in the horizontal direction = 175.13 N - 55.9 N = 119.23 N
6. Calculate the acceleration of the box:
Using Newton's second law of motion, we can determine the acceleration of the box:
Net force = mass × acceleration
Given that the mass of the box is m = 57.0 kg and the net force in the horizontal direction is 119.23 N, we can calculate the acceleration:
Acceleration = Net force / mass
Acceleration = 119.23 N / 57.0 kg = 2.09 m/s²
7. Calculate the work done by the applied force:
The work done by a force can be calculated using the formula:
Work = force × distance × cos(angle)
Given that the applied force is FA = 202 N, the distance is d = 79.0 m, and the angle below the horizontal is 30.0°, we can calculate the work done:
Work = 202 N × 79.0 m × cos(30.0°) = 13,377 J
Therefore, the work done by the applied force is 13,377 J.