The force required to slide the trunk down the inclined plane with constant velocity is approximately F≈75.62N, and the direction of the force is up the incline.
To find the force required to exert on a trunk to make it slide down an inclined plane with constant velocity, we need to consider the forces involved. The forces acting on the trunk are the gravitational force (mg), the normal force (N), the force of friction (f), and the force applied parallel to the incline (F).
Let's denote:
m as the mass of the trunk,
g as the acceleration due to gravity (9.8m/s^2 ),
θ as the angle of the incline (19 degrees),
N as the normal force,
f as the force of friction,
F as the force applied parallel to the incline.
The gravitational force acting parallel to the incline is given by mgsin(θ), and the normal force is mgcos(θ).
The force of friction (f) can be calculated using the formula f=μN, where μ is the coefficient of friction.
Since the trunk is sliding with constant velocity, the net force acting parallel to the incline (F−f) must be zero.
Let's set up the equation:
F−f=0
F−μN=0
F−μmgcos(θ)=0
Now, we can plug in the values and solve for F:
F=μmgcos(θ)
F=(0.322)⋅(m)⋅(9.8)⋅cos(19
F≈75.62N
Once you have the value for F, it represents the force you need to exert on the trunk to make it slide down the incline with constant velocity. The direction of this force will be parallel to the incline.