A 45kg block slides down an incline that is angled at 41 degrees. If the coefficient of kinetic friction between the block and the incline is 0.45 what is the block's acceleration?

normal component of weight = 45*9.81* cos 41

so friction force = 0.45* 45*9.81*cos 41

component of weight down slope = 45*9.81 * sin 41

so
45 a = 45*9.81*sin 41 - .45*45*9.81*cos 41
or
a = 9.81* (sin 41 - .45 cos 41)

The distance from the earth to the sun is 1.5 times 10 to the 11 power m ( 93 million miles) and the time it takes for one complete orbit of the earth about the sun is one year. How long would it take for a planet located at twice this distance from the sun to complete one orbit?

To find the block's acceleration down the incline, we need to use the following formula:

\[a = g \times \sin(\theta) - \mu \times g \times \cos(\theta)\]

where:
- \(a\) is the acceleration
- \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\))
- \(\theta\) is the angle of the incline (\(41^\circ\))
- \(\mu\) is the coefficient of kinetic friction (\(0.45\))

Now, we can substitute these values into the formula:

\[a = 9.8 \, \text{m/s}^2 \times \sin(41^\circ) - 0.45 \times 9.8 \, \text{m/s}^2 \times \cos(41^\circ)\]

Calculating the values, we have:

\[a = 9.8 \, \text{m/s}^2 \times 0.656059 - 0.45 \times 9.8 \, \text{m/s}^2 \times 0.75471\]
\[a = 6.43713 \, \text{m/s}^2 - 3.34815 \, \text{m/s}^2\]
\[a = 3.08898 \, \text{m/s}^2\]

Therefore, the block's acceleration down the incline is approximately \(3.089 \, \text{m/s}^2\).

To find the block's acceleration, you can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration (F = ma).

First, let's calculate the force of gravity acting on the block. The force of gravity can be calculated using the formula F_gravity = m * g, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).

F_gravity = 45 kg * 9.8 m/s^2
F_gravity = 441 N

Next, let's calculate the force of kinetic friction acting on the block. The formula for the force of friction is F_friction = coefficient of friction * normal force. The normal force is the force exerted by a surface perpendicular to the surface. In this case, the normal force is acting in the opposite direction of the force of gravity and can be calculated using the formula F_normal = m * g * cos(theta), where theta is the angle of the incline (41 degrees).

F_normal = 45 kg * 9.8 m/s^2 * cos(41 degrees)
F_normal = 317.36 N

Now, we can calculate the force of kinetic friction:

F_friction = coefficient of friction * F_normal
F_friction = 0.45 * 317.36 N
F_friction = 142.81 N

Since the block is sliding down the incline, the force of kinetic friction will oppose the motion and act in the opposite direction. Therefore, the net force acting on the block can be calculated as:

F_net = F_gravity - F_friction
F_net = 441 N - 142.81 N
F_net = 298.19 N

Now, we can use Newton's second law of motion to find the acceleration:

F_net = ma
298.19 N = 45 kg * a

Solving for a:

a = 298.19 N / 45 kg
a = 6.62 m/s^2

Therefore, the block's acceleration is approximately 6.62 m/s^2.