A crate of mass m (=40 kg) is placed at rest on a (frictionless) inclined plane, which has an angle (= 30o) above horizontal.

a.) The force of gravity on the crate is:

b.) The normal force on the crate is:
size: N

c.) The net force on the crate is:
size: N
direction:

d.) How long would it take the crate to slide 3 m down the incline?
s

e.) If the crate were kept from sliding by a rope pulling parallel to the incline, what would the size of the tension be?
N

a.) The force of gravity on the crate can be found using the equation: F = m * g, where F is the force of gravity, m is the mass of the crate, and g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, F = 40 kg * 9.8 m/s² = 392 N.

b.) The normal force on the crate can be found using the equation: N = m * g * cos(θ), where N is the normal force, m is the mass of the crate, g is the acceleration due to gravity, and θ is the angle of the inclined plane (30° in this case). Therefore, N = 40 kg * 9.8 m/s² * cos(30°) ≈ 333.6 N.

c.) The net force on the crate can be found by calculating the component of the force of gravity that acts parallel to the inclined plane. The force parallel to the plane is given by F_par = m * g * sin(θ), where F_par is the parallel component, m is the mass of the crate, g is the acceleration due to gravity, and θ is the angle of the inclined plane (30° in this case). Therefore, F_par = 40 kg * 9.8 m/s² * sin(30°) ≈ 196 N. Since there are no other forces acting on the crate in the parallel direction, the net force is equal to the force parallel to the plane, which is 196 N.

d.) To find how long it would take for the crate to slide 3 m down the incline, we need to use the equations of motion. The equation that relates displacement, initial velocity, time, and acceleration is: s = ut + (1/2)at^2, where s is the displacement, u is the initial velocity (0 m/s in this case as the crate is initially at rest), a is the acceleration (which can be calculated using a = g * sin(θ) as the crate will accelerate down the incline due to gravity), and t is the time. Rearranging the equation, we get t = sqrt(2s/a). Plugging in the given values, t = sqrt((2 * 3 m) / (9.8 m/s² * sin(30°))) ≈ 1.49 s.

e.) If the crate were kept from sliding by a rope pulling parallel to the incline, the size of the tension in the rope would be equal to the force parallel to the incline, which we calculated earlier as 196 N.

a.) The force of gravity on the crate is given by the equation:

Force of gravity = mass * gravitational acceleration

Using the given mass (m = 40 kg) and assuming the gravitational acceleration is 9.8 m/s^2:

Force of gravity = 40 kg * 9.8 m/s^2 = 392 N

b.) The normal force on the crate can be determined by resolving the gravitational force into its components parallel and perpendicular to the plane. The normal force is equal in magnitude but opposite in direction to the perpendicular component of the gravitational force.

Normal force = perpendicular component of the gravitational force = mg * cos(Θ)

Using the given mass (m = 40 kg) and angle of the plane (Θ = 30°):

Normal force = 40 kg * 9.8 m/s^2 * cos(30°) = 339.4 N

c.) The net force on the crate is the component of the gravitational force parallel to the plane. It can be calculated using the equation:

Net force = mg * sin(Θ)

Using the given mass (m = 40 kg) and angle of the plane (Θ = 30°):

Net force = 40 kg * 9.8 m/s^2 * sin(30°) = 196 N (down the incline)

d.) To calculate the time it would take for the crate to slide 3 m down the incline, we need to know the acceleration of the crate. The acceleration can be determined using the following equation:

Acceleration = Net force / mass

Using the calculated net force (196 N) and given mass (40 kg):

Acceleration = 196 N / 40 kg = 4.9 m/s^2 (down the incline)

To find the time (t), we can use the following kinematic equation for motion on an inclined plane:

Distance = (1/2) * acceleration * time^2

Rearranging the equation to solve for time:

time = sqrt((2 * distance) / acceleration)

Using the given distance (3 m) and calculated acceleration (4.9 m/s^2):

time = sqrt((2 * 3 m) / 4.9 m/s^2) = 1.09 s

e.) If the crate were kept from sliding by a rope pulling parallel to the incline, the tension in the rope would be equal to the component of the gravitational force parallel to the plane. It can be calculated using the equation:

Tension = mg * sin(Θ)

Using the given mass (m = 40 kg) and angle of the plane (Θ = 30°):

Tension = 40 kg * 9.8 m/s^2 * sin(30°) = 196 N