A 5.0-kg box is pulled horizontally with a 45-N force that makes angle of 40 degree with the horizontal. The coefficient of kinetic friction of the surface is 0.20. What is the magnitude of the acceleration of the box?
M*g = 5 * 9.8 = 49 N. = Wt. of box.
Fn = 49 - 45*sin40 = 20.1 N. = Normal
force.
Fk = u*Fn = 0.2 * 20.1 = 4.01 N. = Force
of kinetic friction.
a = (Fx-Fk)/m = (45*Cos40-4.01)/5 = 6.1
m/s^2
To find the magnitude of the acceleration of the box, we need to use Newton's Second Law of Motion, which states that the acceleration of an object is equal to the net force acting on it divided by its mass.
In this case, the net force is the horizontal component of the applied force minus the force of kinetic friction.
1. Find the horizontal component of the applied force:
- Given: Force applied (F) = 45 N
Angle (θ) = 40 degrees
- To find the horizontal component, use the formula:
Horizontal component (F_horizontal) = F * cos(θ)
F_horizontal = 45 N * cos(40°)
2. Find the force of kinetic friction:
- Given: Coefficient of kinetic friction (μ) = 0.20
Normal force (N) = mass * acceleration due to gravity
- To find the normal force, use the formula:
Normal force (N) = mass * acceleration due to gravity
N = 5.0 kg * 9.8 m/s^2
- To find the force of kinetic friction, use the formula:
Force of kinetic friction (F_friction) = μ * N
F_friction = 0.20 * (5.0 kg * 9.8 m/s^2)
3. Calculate the net force:
- Net force (F_net) = F_horizontal - F_friction
4. Calculate the acceleration:
- Acceleration (a) = F_net / mass
Plug the values into the formulas to find the magnitude of the acceleration of the box.