Finding two locations where power out is collision problem

[Samuelb88]

Homework Statement

A small object is shot into a block that is attached to an ideal spring. After the collision, the block moves over a horizontal surface for which it has friction. The masses of the two objects are given as well as the spring constant of the spring k, the initial velocity of the small object and the coeff. of kinetic friction between the block and the surface.

m (small) = .250 kg
M (big) = 3.75 kg
k = 180Nm^-1
v=145m/s
u_k = .600

(B) Determine two locations of the block (assume that it moves back to the left after it has reached its point of maximum spring compression) when the power output of the spring is 120. Watts.

Homework Equations

Power p = (kx)(v)

The Attempt at a Solution

This was part b of the problem. Part a was to find the max compression of the spring which is x_c (max comp) = 1.23 m.power = 120 W = kx * vx

I used the work-energy theorem to find an expression for the velocity from x_c (max comp) to the point x where the power = 120 W. I ended up with this equation

$$K_2 = k \int_{x_c}^{x} x dx - f_kx$$

$$= v = (45x^2-11.75x-68.05))^(^1^/^2^)$$

I substituted the value of v into the expression for power which gives

$$120 = 180x(45x^2-11.75x-68.05))^(^1^/^2^)$$

squaring both sides, i solve for x and get

x_1 = .59m
x_2 = 1.36
x_3 = negative value.all of which are not correct.

my professor provides one answer and that is the power output of the spring is 120 W at x = .091 m.

i don't quite understand what I'm doing wrong.

 

[willem2]

There's a problem with $f_k x$ in your equation for $K_2$ The work done by fricton should be 0 when $x = x_c$

 

[kerimek]

I would approach this problem by first understanding the physical principles involved. In this case, we are dealing with a collision between a small object and a larger block attached to a spring. The key concept here is conservation of energy, which states that the total energy of a system remains constant in the absence of external forces.

In this problem, we are given the masses of the two objects, the spring constant, and the initial velocity of the small object. We are also given the coefficient of kinetic friction, which tells us about the frictional force acting on the block as it moves over the surface. The power output of the spring is related to the work done by the spring, which in turn is related to the change in kinetic energy of the block.

To find the locations where the power output of the spring is 120 Watts, we first need to find the velocity of the block at those locations. This can be done by using the work-energy theorem, as you did in your attempt at a solution. However, it is important to note that the work done by the spring is equal to the change in kinetic energy of the block only in the absence of external forces. In this case, there is also a frictional force acting on the block, which means that the work done by the spring is not equal to the change in kinetic energy of the block.

To account for the frictional force, we can use the work-energy theorem again, this time taking into account the work done by friction. This will give us a more accurate expression for the velocity of the block at the locations where the power output of the spring is 120 Watts.

Once we have the velocity, we can use the equation for power (P = Fv) to find the force acting on the block at those locations. This force will be equal to the force exerted by the spring, since the frictional force does not contribute to the power output of the spring.

Using the force and the given mass of the block, we can solve for the distance x traveled by the block, which will give us the two locations where the power output of the spring is 120 Watts.

In summary, to solve this problem correctly, we need to take into account the frictional force acting on the block and use the work-energy theorem twice to find the velocity and distance traveled by the block at the locations where the power output of the spring is 120 Watts.