A crate of mass m(=32 kg) rides on the bed of a truck attached by a cord to the back of the cab . The cord can withstand a maximum tension of 68N before breaking. Neglecting friction between the crate and the truck bed, find the maximum acceleration the truck can have before the cord breaks.

a = F/M = 68/32 = 2.125 m/s^2.

a=f/m

68/32= 2.12 m/s^2

Well, this sounds like a tricky situation. It seems the crate wants to have some fun and ride on the truck, but the cord has other plans. Let's see if we can help them find some common ground.

To find the maximum acceleration the truck can have before the cord breaks, we need to consider the forces acting on the crate. Since the cord is attached to the back of the cab, there are two forces acting on the crate: the tension in the cord, and the weight of the crate.

The weight of the crate can be calculated using the equation W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. So in this case, W = (32 kg) × (9.8 m/s²) = 313.6 N.

Now we know the maximum tension the cord can withstand is 68 N. So, we equate the tension in the cord to the weight of the crate and solve for the maximum acceleration:

68 N = 313.6 N - m × a

Since we're looking for the maximum acceleration, we want to find the minimum value for the mass of the crate. So, we can use the maximum value for the tension, plug in the given values, and solve for a:

68 N = 313.6 N - (32 kg) × a

Rearranging the equation, we find:

(32 kg) × a = 313.6 N - 68 N

(32 kg) × a = 245.6 N

Dividing through by 32 kg, we get:

a = 245.6 N / 32 kg

a ≈ 7.675 m/s²

So, the maximum acceleration the truck can have before the cord breaks is approximately 7.675 m/s². Just make sure not to accelerate too fast, or the crate might become an unexpected stowaway!

To find the maximum acceleration the truck can have before the cord breaks, we need to consider the forces acting on the crate.

1. Gravitational force:
The weight of the crate is given by the equation F_gravity = m * g, where m is the mass of the crate (32 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). Plugging in the values, we get F_gravity = 32 kg * 9.8 m/s^2 = 313.6 N.

2. Tension force:
The tension force in the cord is equal to the force required to accelerate the crate. Let's call this force F_tension. Since the crate is attached to the truck and is accelerating along with it, the acceleration of the crate is the same as the acceleration of the truck.

Given that the tension force must not exceed 68 N, we have F_tension ≤ 68 N.

Since the maximum acceleration is our unknown, let's call it a_max. Therefore, we have F_tension = m * a_max.

Combining the above equations, we can write the equation as m * a_max ≤ 68 N.

Plugging in the mass of the crate, we have 32 kg * a_max ≤ 68 N.

Now, we can solve for a_max by dividing both sides of the inequality by 32 kg:

a_max ≤ 68 N / 32 kg.

Calculating this expression, we find:

a_max ≤ 2.125 m/s^2.

Therefore, the maximum acceleration the truck can have before the cord breaks is 2.125 m/s^2.

To find the maximum acceleration the truck can have before the cord breaks, we need to consider the forces acting on the crate.

First, we need to determine the force acting on the crate. There are two forces acting on the crate: the tension in the cord pulling the crate forward and the weight of the crate pulling it downward.

The weight of the crate is given by the formula:

Weight = mass x gravitational acceleration

Plugging in the values:

Weight = 32 kg x 9.8 m/s² (gravitational acceleration)

Weight = 313.6 N

Since the crate is in equilibrium, the tension in the cord is equal to the weight of the crate (assuming no friction). Thus, the tension in the cord is 313.6 N.

Now, let's consider the maximum tension the cord can withstand before breaking, which is given as 68 N.

To solve for the maximum acceleration, we can set up the following equation:

Tension in the cord ≤ Maximum tension

313.6 N ≤ 68 N

Since the tension in the cord exceeds the maximum tension, the cord will break if the crate experiences an acceleration greater than the maximum acceleration it can withstand.

Thus, the maximum acceleration the truck can have before the cord breaks is 0 m/s². The crate will stay stationary as long as the tension in the cord does not exceed the maximum tension of 68 N.