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
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,
T= m2•a - F= 7.5•2.9 – 26.1 = 4.35 N.

A train moving with speed 144 km/ hr stop is 8 sec .find the distance travelled before stopping

Question #1: To find the maximum acceleration a car can undergo, we need to use the coefficient of static friction between the tires and the ground. The maximum acceleration can be calculated using the formula:

a_max = μ_s * g

where μ_s is the coefficient of static friction and g is the acceleration due to gravity, approximately 9.8 m/s^2.

To find the answer, simply multiply the coefficient of static friction (0.82) by the acceleration due to gravity (9.8 m/s^2):

a_max = 0.82 * 9.8 = 8.036 m/s^2

Therefore, the maximum acceleration a car can undergo with a coefficient of static friction of 0.82 is approximately 8.036 m/s^2.

Question #2: To answer both parts of the question, we need to understand the concept of free body diagrams and Newton's second law.

Part a) Newton's second law states that the force exerted on an object is equal to its mass multiplied by its acceleration. In this scenario, the force F applied to the bottom block is causing the top block to accelerate to the right. Assuming the acceleration of the bottom block is the same, we can set up the following equation:

F - m_top * a = m_bottom * a

Where F is the force applied, m_top is the mass of the top block, m_bottom is the mass of the bottom block, and a is the acceleration of both blocks.

Rearranging the equation, we get:

F = (m_top + m_bottom) * a

Substituting the given values, we have:

F = (1.5 kg + 7.5 kg) * 2.9 m/s^2

F = 9 kg * 2.9 m/s^2

F = 26.1 N

Therefore, the force F that must be applied to the bottom block so that the top block accelerates to the right at 2.9 m/s^2 is 26.1 N.

Part b) To find the tension in the connecting cord, we need to understand that the tension is the same throughout the cord when there is no significant friction. Assuming the tension is T, we can set up an equation for the forces acting on the top block:

T - m_top * g = m_top * a

Simplifying the equation, we get:

T = m_top * (g + a)

Substituting the given values, we have:

T = 1.5 kg * (9.8 m/s^2 + 2.9 m/s^2)

T = 1.5 kg * 12.7 m/s^2

T = 19.05 N

Therefore, the tension in the connecting cord is 19.05 N.