021 (part 1 of 3) 10.0 points A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are µs = 0.62 and µk = 0.53,

respectively. The acceleration of gravity is 9.8 m/s2 .
36 kg µ
29◦
What is the frictional force acting on the 36 kg mass? Answer in units of N.
022 (part 2 of 3) 10.0 points What is the largest angle which the incline can have so that the mass does not slide down the incline? Answer in units of ◦.
023 (part 3 of 3) 10.0 points What is the acceleration of the block down the incline if the angle of the incline is 36◦ ? Answer in units of m/s2.

Oh, it's time for some physics fun! Let's tackle these questions one by one!

021: To find the frictional force acting on the 36 kg mass, we first need to calculate the maximum static frictional force. We can use the formula F_friction = µ_s * F_normal, where F_normal is the normal force acting on the object.

Since the object is at rest, the normal force is equal to the gravitational force acting on it. So, F_normal = m * g, where m is the mass and g is the acceleration due to gravity.

F_normal = 36 kg * 9.8 m/s² = 352.8 N.

Now, let's calculate the maximum static frictional force:

F_friction_max = µ_s * F_normal = 0.62 * 352.8 N ≈ 218.54 N.

So, the frictional force acting on the 36 kg mass is approximately 218.54 N.

022: To find the largest angle at which the mass does not slide down the incline, we need to find the angle at which the gravitational force component parallel to the incline is equal to the maximum static frictional force.

Let's set up the equation: m * g * sin(θ) = µ_s * m * g * cos(θ), where θ is the angle of the incline.

Since the mass cancels out, we can simplify the equation: sin(θ) = µ_s * cos(θ).

To solve this, we can take the inverse tangent of both sides of the equation: θ = atan(µ_s).

θ = atan(0.62) ≈ 31.3°.

So, the largest angle at which the mass does not slide down the incline is approximately 31.3°.

023: Finally, to find the acceleration of the block down the incline when the angle is 36°, we need to calculate the net force acting on the block.

The net force is given by the difference between the gravitational force component parallel to the incline and the kinetic frictional force.

F_net = m * g * sin(θ) - µ_k * m * g * cos(θ).

Let's substitute the values: F_net = 36 kg * 9.8 m/s² * sin(36°) - 0.53 * 36 kg * 9.8 m/s² * cos(36°).

Calculating this will give us the net force, which we can then use to find the acceleration using Newton's second law, F = m * a.

But I'll let you do the calculations this time! Keep up the good work and remember, physics can be fun when you don't slide down the slippery slope of boredom!

For part 1 of the question:

The frictional force can be calculated using the equation:

Frictional force = µs * Normal force

The normal force can be calculated using the equation:

Normal force = Mass * gravitational acceleration * cos(theta)

Substituting the values given:

Mass = 36 kg
Gravitational acceleration = 9.8 m/s^2
Theta = 29°
µs = 0.62

Normal force = 36 kg * 9.8 m/s^2 * cos(29°)

Now, calculate the frictional force:

Frictional force = 0.62 * (36 kg * 9.8 m/s^2 * cos(29°))

Solve for the frictional force to get the answer in N.

To find the answers to these questions, we will use the laws of friction and Newton's second law of motion.

For question 021, we need to find the frictional force acting on the 36 kg mass. The frictional force is given by the equation:

Frictional force = coefficient of friction * normal force

The normal force can be calculated as the component of the weight of the mass perpendicular to the incline, which is:

Normal force = mass * acceleration due to gravity * cos(angle)

Plugging in the given values, we have:

Normal force = 36 kg * 9.8 m/s^2 * cos(29°)

Now, we can substitute the value of the coefficient of static friction (µs = 0.62) into the formula to find the frictional force:

Frictional force = µs * Normal force

Frictional force = 0.62 * (36 kg * 9.8 m/s^2 * cos(29°))

Calculating this value will give us the answer to question 021 in units of Newtons (N).

For question 022, we need to find the largest angle at which the mass does not slide down the incline. In this case, the force of friction exactly balances the component of the weight parallel to the incline, preventing the mass from sliding. The equation for this is:

Frictional force = coefficient of static friction * Normal force

Setting the frictional force equal to the weight component parallel to the incline, we have:

µs * Normal force = mass * acceleration due to gravity * sin(angle)

Solving for the angle, we get:

angle = arcsin((µs * Normal force) / (mass * acceleration due to gravity))

Substituting the given values, we can calculate the angle and provide the answer in units of degrees (°).

For question 023, we need to find the acceleration of the block down the incline when the angle is 36°. To find this, we can use Newton's second law of motion:

Sum of forces = mass * acceleration

The forces acting on the block are the gravitational force component parallel to the incline (weight * sin(angle)) and the frictional force (µk * Normal force). The sum of these forces is equal to the mass times acceleration. Solving for the acceleration, we have:

acceleration = (weight * sin(angle) - µk * Normal force) / mass

Substituting the given values, we can calculate the acceleration and provide the answer in units of m/s^2.

M*g = 36 * 9.8 = 352.8 N. = Wt. of block.

Fp = 352.8*sin29o = 171 N. = Force parallel to the incline.

Fn = 352.8*Cos29 = 308.6 N. = Normal force.

1. Fs = u*Fn = 0.62 * 308.6 = 191.3 N. = Force of static friction.

2. Fp = Fs.
352.8*sin A = 191.3
A = ?.

3. Fp = 352.8*sin36 = 207.4 N.

Fn = 352.8*Cos36 = 285.4 N.

Fk = u*Fn = 0.53 * 285.4 N. = 151.3 N. = Force of kinetic friction.

Fp-Fk = M*a.
207.4-151.3 = 36*a
a = ?.