A 100 kg crate rest at the back of the bed of a pick up truck. The length of the crate is 1 m in the length of the bed is 4 m. There is no friction between the bed of the truck in the crate. If the truck is traveling at 25 m/s just before it hits the wall, how long before the crate hits the back of the pickup's cab? How would the answer change if the crates mass were 1000 kg?

geaghaeghaerf

To calculate the time it takes for the crate to hit the back of the pickup's cab, we need to determine the acceleration of the crate and then use the equation of motion to find the time.

Given:
Mass of the crate (m) = 100 kg
Length of the crate (l) = 1 m
Length of the truck bed (L) = 4 m
Truck's velocity (v) = 25 m/s

First, we need to calculate the acceleration of the crate. Since there is no friction between the bed of the truck and the crate, the only force acting on the crate is the force due to the truck's acceleration. This force can be calculated using Newton's second law:

Force (F) = mass (m) * acceleration (a)

In this case, the force is acting horizontally in the direction of the truck's motion. The force is given by:

Force (F) = mass of the crate (m) * acceleration of the crate (a)

Since the crate is at rest on the bed of the pickup truck, the net force on the crate is zero. Therefore, the force due to the truck's acceleration is equal in magnitude and opposite in direction to the force due to static friction.

The force of static friction can be calculated using the equation:

Force of static friction (fs) = coefficient of static friction (μs) * normal force (N)

The normal force (N) is equal to the weight of the crate (mg), where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Therefore, fs = μs * mg

The force of static friction (fs) is equal to the force due to the truck's acceleration (F). Hence, we can equate the two equations:

μs * mg = m * a

Simplifying the equation by canceling the mass of the crate (m) on both sides:

μs * g = a

Now, we have the acceleration of the crate (a) as μs * g.

Now, we can use the equation of motion to calculate the time it takes for the crate to hit the back of the pickup's cab. The equation of motion used here is:

Distance (d) = Initial velocity (v) * Time (t) + (1/2) * Acceleration (a) * Time^2

In this case, the initial velocity is the truck's velocity (v), the distance travelled is from the back of the truck bed to the back of the cab (L-l), and the acceleration is the acceleration of the crate (a).

Substituting the given values into the equation, we get:

L - l = v * t + (1/2) * a * t^2

Rearranging the equation to solve for time (t):

(1/2) * a * t^2 + v * t + (l - L) = 0

We can now substitute the value of acceleration (a) we calculated earlier and solve the quadratic equation to find the time it takes for the crate to hit the back of the pickup's cab.

Now, let's solve this equation for both scenarios, first for a crate mass of 100 kg and then for a crate mass of 1000 kg.

For a 100 kg crate:

μs * g = a = (coefficient of static friction) * (acceleration due to gravity)

Assuming a coefficient of static friction of 0.3:

a = 0.3 * 9.8 ≈ 2.94 m/s^2

Substituting this value into the equation:

(1/2) * 2.94 * t^2 + 25 * t + (1 - 4) = 0

Simplifying the equation:

1.47 * t^2 + 25 * t - 3 = 0

Using the quadratic formula:

t ≈ 0.070 s (rounded to 3 decimal places)

For a 1000 kg crate:

μs * g = a = (coefficient of static friction) * (acceleration due to gravity)

Assuming a coefficient of static friction of 0.3:

a = 0.3 * 9.8 ≈ 2.94 m/s^2

Substituting this value into the equation:

(1/2) * 2.94 * t^2 + 25 * t + (1 - 4) = 0

Simplifying the equation:

1.47 * t^2 + 25 * t - 3 = 0

Using the quadratic formula:

t ≈ -0.351 s or 1.988 s

Since time cannot be negative in this context, the time it takes for the 1000 kg crate to hit the back of the pickup's cab is approximately 1.988 seconds.

Therefore, with a crate mass of 100 kg, it takes about 0.070 seconds for it to hit the back of the pickup's cab, while with a crate mass of 1000 kg, it takes approximately 1.988 seconds.