Divide the net horizontal forward force by the mass.
[173cos27.5 -(M*g -173sin27.5)*ìk]/M = a
M = 27 kg
ìk = 0.2
g = 9.81
Solve for a.
by pulling on a rope attached to the box. The
student pulls with a force of 173 N at an angle of 27.5� above the horizontal. The box has a
mass of 27 kg, and μk between the box and
the floor is 0.2.
The acceleration of gravity is 9.81 m/s2 .
Find the acceleration of the box.
Answer in units of m/s2
[173cos27.5 -(M*g -173sin27.5)*ìk]/M = a
M = 27 kg
ìk = 0.2
g = 9.81
Solve for a.
The component of force parallel to the floor is given by:
F_parallel = F * cos(angle)
F_parallel = 173 N * cos(27.5º)
F_parallel ≈ 156.16 N
The force of friction acting on the box can be calculated using the equation:
Frictional force = coefficient of friction * normal force
Normal force = mass * gravitational acceleration
Normal force = 27 kg * 9.81 m/s^2
Normal force ≈ 264.87 N
Frictional force = 0.2 * 264.87 N
Frictional force ≈ 52.97 N
Since the box is moving, the applied force F_parallel must be greater than the force of friction. Therefore:
F_parallel > Frictional force
We can use Newton's second law to calculate the acceleration of the box:
F_parallel - Frictional force = mass * acceleration
156.16 N - 52.97 N = 27 kg * acceleration
103.19 N = 27 kg * acceleration
Acceleration = 103.19 N / 27 kg
Acceleration ≈ 3.82 m/s²
Therefore, the acceleration of the box is approximately 3.82 m/s².
First, we need to determine the horizontal and vertical components of the force pulling the box. The horizontal component can be found using the formula: Fx = F * cos(theta), where F is the applied force and theta is the angle above the horizontal.
Fx = 173 N * cos(27.5 degrees) = 155.25 N
The frictional force opposing the motion of the box can be found using the formula: Ff = μk * N, where μk is the coefficient of kinetic friction and N is the normal force.
The normal force N is equal to the weight of the box, which can be found using the formula: N = m * g, where m is the mass of the box and g is the acceleration due to gravity.
N = 27 kg * 9.81 m/s^2 = 264.87 N
Ff = 0.2 * 264.87 N = 52.97 N
Since the box is moving horizontally, the net force in the horizontal direction is equal to the applied force minus the frictional force.
Net force in the horizontal direction = Fx - Ff = 155.25 N - 52.97 N = 102.28 N
Finally, we can find the acceleration of the box using Newton's second law:
Net force = mass * acceleration
102.28 N = 27 kg * acceleration
acceleration = 102.28 N / 27 kg ≈ 3.788 m/s^2
Therefore, the acceleration of the box is approximately 3.788 m/s^2.