Question

You give a crate of mass m=1.00kg has an initial speed of vi=4.00m/s up an incline of θ=30.7∘ θ=30.7∘. the crate then slides along the incline, reaches a spring of spring constant k=45.4n/m , and compresses the spring by δs=15.0cm before stopping.assuming there is no friction, what is the distance, d, in meters along the incline that the crate slides.

Answer 1

The distance the crate slides along the incline is approximately 0.295 meters, determined by using conservation of energy and basic trigonometry.

Step-by-step explanation:

To calculate the distance along the incline that a crate of mass m slides before compressing a spring, we can use conservation of energy principles. As there is no friction, the mechanical energy of the system is conserved.

The crate starts with initial kinetic energy due to its velocity and loses this energy as it compresses the spring. Thus, the initial kinetic energy of the crate is equal to the potential energy stored in the spring at maximum compression:

Kinetic Energy = ½mv²
Potential Energy = ½kδs²

Setting these equal and solving for the initial velocity:

½(1.00kg)(4.00m/s)² = ½(45.4N/m)(0.15m)²

Now, using trigonometry, we can find the distance along the incline, d. If θ is the angle of the incline:

d = δs / sin(θ)

Substituting the given values and solving:

d = 0.15m / sin(30.7°) = 0.15m / 0.5075 = 0.295m

So the distance, d, the crate slides along the incline is approximately 0.295 meters.