A 10-kg box is pulled along a horizontal surface by a
force of 40N applied at a 30° angle above
horizontal. The coefficient of kinetic friction is 0.3.
Calculate the acceleration
Wb = M*g = 10*9.8 = 98 N.
Fp = 98*sin 0 = 0 = Force parallel to the surface.
Fn = 98*Cos 0 - 40*sin30 = 78 N. = Normal force.
Fk = u*Fn = 0.3 * 78 = 23.4 N. = Force of kinetic friction.
a=(Fap*Cos30-Fp-Fk)/M=(34.6-0-23.4)/10 = 1.12 m/s^2.
Find Acceleration if coefficient of friction is 0.2. and applied force is 39.6 .
Well, well, well! Looks like the box is getting a little pushy, huh? Let's find out how fast it's gonna zoom!
First, we need to find the force of friction. The formula for frictional force is: frictional force = coefficient of kinetic friction * normal force. The normal force is equal to the weight of the box, which is mass * acceleration due to gravity.
So, the normal force = 10 kg * 9.8 m/s^2 = 98 N.
Now, let's calculate the force of friction = 0.3 * 98 N = 29.4 N.
Since the force applied is at an angle above the horizontal, we need to find its horizontal component. The horizontal component of the force = force applied * cos(angle).
So, horizontal force = 40 N * cos(30°) = 40 N * 0.866 = 34.64 N.
Now, let's calculate the net force acting on the box. Net force = horizontal force - force of friction.
Net force = 34.64 N - 29.4 N = 5.24 N.
Finally, we can calculate the acceleration using Newton's second law. Net force = mass * acceleration.
Acceleration = net force / mass = 5.24 N / 10 kg = 0.524 m/s^2.
Phew! The box is gonna zoom away with an acceleration of 0.524 m/s^2. Watch out, world!
To calculate the acceleration of the box, we need to consider the forces acting on it. In this case, we have three main forces: the applied force, the force of gravity, and the force of friction.
First, let's break down the applied force into its horizontal and vertical components. The horizontal component can be calculated using the formula:
F_horizontal = F_applied * cos(angle)
where:
F_applied = 40N (applied force)
angle = 30° (angle above horizontal)
So, the horizontal component of the applied force is:
F_horizontal = 40N * cos(30°) = 40N * 0.866 = 34.64N
Next, let's evaluate the force of gravity acting on the box. The force of gravity can be calculated using the formula:
F_gravity = m * g
where:
m = 10kg (mass of the box)
g = 9.8 m/s^2 (acceleration due to gravity)
So, the force of gravity is:
F_gravity = 10kg * 9.8 m/s^2 = 98N
Now, let's calculate the force of friction. The force of friction can be determined using the formula:
F_friction = μ * N
where:
μ = 0.3 (coefficient of kinetic friction)
N = F_normal (normal force)
The normal force can be found as the vertical component of the force of gravity when the box is on a horizontal surface, which is equal to the weight of the box:
F_normal = F_gravity = 98N
So, the force of friction is:
F_friction = 0.3 * 98N = 29.4N
Now, we can calculate the net force acting on the box in the horizontal direction. The net force is the vector sum of the applied and frictional forces, taking into account their directions:
Net Force = F_horizontal - F_friction
Net Force = 34.64N - 29.4N = 5.24N
Finally, to calculate the acceleration, we can use Newton's second law of motion, which states that the acceleration is equal to the net force divided by the mass:
a = Net Force / m
a = 5.24N / 10kg = 0.524 m/s^2
Therefore, the acceleration of the box is 0.524 m/s^2.