1.A student moves a box of books down the hall
by pulling on a rope attached to the box. The
student pulls with a force of 181 N at an angle
of 20.8◦ above the horizontal. The box has a
mass of 40.3 kg, and μk between the box and
the floor is 0.29.
The acceleration of gravity is 9.81 m/s2 .
Find the acceleration of the box.
2.Now the student moves the box up a ramp
(with the same coefficient of friction) inclined
at 10.3◦ with the horizontal.
b) If the box starts from rest at the bottom
of the ramp and is pulled at an angle of 20.8◦
with respect to the incline and with the same
181 N force, what is the acceleration up the
ramp?
Answer in units of m/s2.
1. Break the pulling force in to a vertical and horizontal component. The vertical component subtracts from the normal weight. So figure friction based on that.
Net horizontal force=m*a
horizonalPulling-friction= ma
sove force.
2. Again, break the pulling into a normal component and a parallel(plane) component.
the normal force will be mg*cosTeta-normalofPullingcomponent.
Friction= mu*normal force above.
Component of gravity down the plane, mgSinTheta..
Net up force=ma
ParallelPulling-friction-gravitydownplane=ma
solve for a
Fx = 181 cos 20.8
normal force = m g - 181 sin 20.8
friction force =.29(mg-181 sin 20.8)
ma= 181cos 20.8-.29(mg-181 sin 20.8)
divide by m to get a
now you do the second one the same way
1. 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 product of its mass and acceleration (F = ma).
First, we need to resolve the pulling force into its horizontal and vertical components. The vertical component can be found by multiplying the force (181 N) by the sine of the angle (20.8°):
Vertical force (Fy) = 181 N * sin(20.8°)
Now, we can calculate the vertical acceleration (ay) using the formula F = ma:
Fy = (mass of box) * ay
181 N * sin(20.8°) = (40.3 kg) * ay
Rearranging the equation, we can solve for ay:
ay = (181 N * sin(20.8°)) / (40.3 kg)
By substituting the given values into the equation, we can calculate the acceleration of the box.
2. Since the box is moving up a ramp inclined at 10.3°, we need to consider the forces acting parallel and perpendicular to the incline.
First, we need to calculate the gravitational force acting on the box. The downward force parallel to the incline is given by:
Force due to gravity (Fg) = (mass of box) * (acceleration due to gravity) * sin(10.3°)
Next, we need to resolve the pulling force into its parallel and perpendicular components. The parallel component can be found by multiplying the force (181 N) by the cosine of the angle (20.8°):
Parallel force (Fp) = 181 N * cos(20.8°)
Using the coefficient of kinetic friction (μk), we can find the frictional force acting on the box:
Frictional force (Ff) = μk * (force due to gravity)
To find the net force acting on the box along the incline, we need to subtract the frictional force from the parallel force:
Net force along the incline (Fnet) = Fp - Ff
Now, we can calculate the acceleration up the ramp using the formula F = ma:
Fnet = (mass of box) * (acceleration up the ramp)
Fp - Ff = (mass of box) * (acceleration up the ramp)
Rearranging the equation, we can solve for the acceleration up the ramp:
acceleration up the ramp = (Fp - Ff) / (mass of box)
By substituting the given values into the equation, we can calculate the acceleration up the ramp.
To find the acceleration of the box in both scenarios, we need to analyze the forces acting on the box and apply Newton's second law of motion.
1. Finding the acceleration of the box when pulled horizontally:
The horizontal component of the applied force is responsible for accelerating the box, overcoming the frictional force and the inertia of the box.
First, calculate the horizontal component of the applied force:
F_horizontal = F_applied * cos(theta)
F_horizontal = 181 N * cos(20.8°)
F_horizontal = 181 N * 0.9363
F_horizontal = 169.6 N
Next, calculate the net force acting on the box in the horizontal direction:
F_net = F_horizontal - frictional force
F_net = F_horizontal - μk * m * g
F_net = 169.6 N - 0.29 * 40.3 kg * 9.81 m/s^2
F_net = 169.6 N - 115.2 N
F_net = 54.4 N
Using Newton's second law, we can determine the acceleration:
F_net = m * a
a = F_net / m
a = 54.4 N / 40.3 kg
a ≈ 1.35 m/s^2
Therefore, the acceleration of the box when pulled horizontally is approximately 1.35 m/s^2.
2. Finding the acceleration of the box when pulled up the inclined ramp:
In this scenario, the applied force has both horizontal and vertical components.
First, calculate the horizontal component of the applied force:
F_horizontal = F_applied * cos(theta)
F_horizontal = 181 N * cos(20.8°)
F_horizontal = 181 N * 0.9363
F_horizontal = 169.6 N
Next, calculate the vertical component of the applied force:
F_vertical = F_applied * sin(theta)
F_vertical = 181 N * sin(20.8°)
F_vertical = 181 N * 0.3420
F_vertical = 62.0 N
Now, resolve the weight of the box into components:
Weight_vertical = m * g * sin(theta)
Weight_vertical = 40.3 kg * 9.81 m/s^2 * sin(10.3°)
Weight_vertical = 40.3 kg * 9.81 m/s^2 * 0.177
Weight_vertical = 70.4 N
Calculate the net force acting on the box parallel to the ramp:
F_parallel = F_horizontal - frictional force
F_parallel = 169.6 N - μk * m * g * cos(theta)
F_parallel = 169.6 N - 0.29 * 40.3 kg * 9.81 m/s^2 * cos(10.3°)
F_parallel = 169.6 N - 112.8 N
F_parallel = 56.8 N
Using Newton's second law, we can determine the acceleration parallel to the ramp:
F_net = m * a_parallel
a_parallel = F_parallel / m
a_parallel = 56.8 N / 40.3 kg
a_parallel ≈ 1.41 m/s^2
Therefore, the acceleration of the box when pulled up the ramp is approximately 1.41 m/s^2.