Question # 1 : What is the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the ground is 0.82?

Question # 2 : A 1.5-kg block rests on top of a 7.5kg block. The cord and pulley have negligible mass, and there is no significant friction anywhere.

Part a) What force F must be applied to the bottom block so the top block accelerates to the right at 2.9 m/s^2 ?

Part b) What is the tension in the connecting cord?

Q1:

The car moves forward by the reaction force from the ground, produced due to friction between tires and road. According to the Newton’s 3rd law
F12 = F21
F(net)max = F(friction)max
m•a = k• m•g
a = k• g = 0.82•9.8 = 8.04 m/s^2
Q2:
m1 =1.5 kg, m2 = 7.5 kg, a=2.9 m/s²
(a) The horizontal projections of the equations of motion for each block are
m1•a = T,
m2•a = T-F,
F = (m1+m2) •a = (1.5+7.5) •2.9 = 26.1 N,
(b) T= m2•a - F= 7.5•2.9 – 26.1 = 4.35 N.

Question #1: The maximum acceleration a car can undergo with a coefficient of static friction of 0.82? Well, let's just say it's so fast that it might break the space-time continuum. You might need a DeLorean and a flux capacitor for that kind of acceleration!

Question #2:
a) To make the top block accelerate to the right at 2.9 m/s^2, you'll need a force that's strong enough to convince the block that it's time to get moving! Maybe you could use a motivational speaker to inspire it. Just be careful not to hire a pushy one.

b) Ah, the tension in the connecting cord. Sounds like a sticky situation! You'll have to pull it tight enough, but not too tight, because you don't want the cord to snap like a bad punchline. Remember, tension is like dealing with your annoying neighbors – it can be a delicate balance!

Question # 1 : To determine the maximum acceleration a car can undergo, we will use the formula for maximum static friction:

f_s = μ_s * N

where f_s is the maximum static friction, μ_s is the coefficient of static friction, and N is the normal force.

Assuming that the normal force is equal to the weight of the car (N = mg), where m is the mass of the car and g is the acceleration due to gravity, we can rewrite the formula as:

f_s = μ_s * m * g

The maximum acceleration (a_max) that the car can undergo is equal to the maximum static friction divided by the mass of the car:

a_max = f_s / m

Substituting the given values, we have:

a_max = (0.82) * (m) * (g) / m

Since the mass of the car cancels out, the maximum acceleration a car can undergo is:

a_max = 0.82 * g

Now, we need to determine the value of acceleration due to gravity (g). On Earth, the value of gravitational acceleration is approximately 9.8 m/s^2.

Substituting this value, we have:

a_max = 0.82 * 9.8

a_max ≈ 8.036 m/s^2

Therefore, the maximum acceleration a car can undergo if the coefficient of static friction between the tires and the ground is 0.82 is approximately 8.036 m/s^2.

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Question # 2 :

Part a) To calculate the force required to accelerate the top block to the right at 2.9 m/s^2, we can use Newton's second law of motion:

F = m * a

where F is the net force applied to the block, m is the mass of the block, and a is the acceleration of the block.

In this case, we have two blocks connected by a cord, and the force applied to the bottom block will be transmitted to the top block. So, the net force acting on the system is the force applied at the bottom block.

Let's consider the forces acting on the bottom block:

- Weight (W_1) = m_1 * g, where m_1 is the mass of the bottom block and g is the acceleration due to gravity.
- Tension in the cord (T).

Since the bottom block is accelerating to the right, we have:

F - T = m_1 * a
F = T + m_1 * a

To accelerate the top block, the force applied at the bottom block must be equal to the sum of the masses of both blocks multiplied by the desired acceleration:

F = (m_1 + m_2) * a

Substituting the given values, we have:

F = (m_1 + m_2) * 2.9

Part b) To calculate the tension in the connecting cord, we can use the force equation again. Since the cord is not stretchable and has no mass, the tension in the cord is the same throughout the cord. Therefore, the tension in the cord is equal to the force applied at the bottom block.

Thus, the tension in the connecting cord is also given by:

Tension = (m_1 + m_2) * 2.9

Please provide the values of m1 and m2 in order to proceed with the calculations.

Question # 1 : To find the maximum acceleration a car can undergo, we need to consider the relationship between the coefficient of static friction and the maximum acceleration.

The maximum acceleration of a car is limited by the maximum static friction force between the tires and the ground. The formula for static friction force is given by F_friction = μ_s * N, where μ_s is the coefficient of static friction and N is the normal force.

Now, technically, the normal force acting on the car is equal to the weight of the car, which is given by W = m * g, where m is the mass of the car and g is the acceleration due to gravity.

To find the maximum acceleration, we need to express the static friction force in terms of acceleration. Rearranging the formula for static friction force, we get F_friction = μ_s * m * g.

Since the force of friction is also equal to the mass of the car multiplied by its acceleration (F_friction = m * a), we can equate the two expressions to find the maximum acceleration:

m * a = μ_s * m * g

Canceling out the mass on both sides of the equation, we get:

a = μ_s * g

Therefore, to find the maximum acceleration a car can undergo, multiply the coefficient of static friction (0.82 in this case) by the acceleration due to gravity (approximately 9.8 m/s^2).

a = 0.82 * 9.8
a ≈ 8.04 m/s^2

Therefore, the maximum acceleration a car can undergo is approximately 8.04 m/s^2.

Question # 2 : Part a) To find the force F needed to accelerate the top block to the right, we need to consider Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a).

In this case, the force F applied to the bottom block causes both blocks to accelerate to the right. Since the two blocks are connected by a cord and pulley with no significant friction, the acceleration of the bottom block will be the same as the top block.

Given that the acceleration of the top block is 2.9 m/s^2 and the mass of the bottom block is 7.5 kg, we can calculate the force F required using Newton's second law:

F = m * a
F = 7.5 kg * 2.9 m/s^2
F ≈ 21.75 N

Therefore, the force F required to accelerate the bottom block so that the top block accelerates to the right at 2.9 m/s^2 is approximately 21.75 N.

Part b) To find the tension in the connecting cord, we need to consider the forces acting on both blocks.

In this scenario, the tension in the connecting cord will be the same for both blocks. This can be explained by applying Newton's second law of motion again. The force applied to the bottom block, F, causes both the bottom and top blocks to accelerate. Since the cord is rigid and has negligible mass, the tension in the cord will be the same throughout.

Therefore, the tension in the connecting cord is equal to the force applied to the bottom block, which we already calculated as 21.75 N in part a).

Therefore, the tension in the connecting cord is approximately 21.75 N.