a 10.0 g bullet is fired into a 1.0 kg block initially at rest on a rough surface and connected to a spring with k = 200 N/m. Immediately following the collision the block has a speed of 2.0 m/s. The coefficient of friction between the ground and the block is 0.10.
a) If the block slides a distance x after the collision before turning around, what is the work done by friction over that distance (in terms of x)?
W = -umgx
= -(0.10)(1.0kg)(9.8m/s^2)x
= -0.98x
b) What is the mechanical energy of the block immediately following the collision?
E_f = .5mv_f^2 + .5kx^2
= .5(1kg)(2.0m/s^2) + .5(200N/m)x^2
= 2kg*m^2/s^2 + (100N/m)x^2
c) What is the mechanical energy of the block when it has reached its maximum compression distance x?
d) How does the value of mechanical energy immediately following the collision compare to the mechanical energy at the maximum compression point of the block? Be specific.
e) Calculate x.
f) What is the maximum velocity reached by the block as it returns to its unstretched position following compression?
To answer part c) of the question, we need to determine the maximum compression distance x. Let's start by finding the initial compression of the spring.
The initial speed of the block after the collision is given as 2.0 m/s. We can equate the kinetic energy at that point to the potential energy stored in the spring:
.5mv_f^2 = .5kx^2
Plugging in the values, we have:
.5(1 kg)(2.0 m/s)^2 = .5(200 N/m)x^2
Simplifying:
2 kg*m^2/s^2 = 100 N/m * x^2
Now, to find the maximum compression distance x, we rearrange the equation:
x^2 = (2 kg*m^2/s^2) / (100 N/m)
x^2 = 0.02 m^2/s^2
x = sqrt(0.02) m
So, the maximum compression of the spring is approximately 0.141 m.
Moving on to part d), we need to compare the mechanical energy immediately following the collision to the mechanical energy at the maximum compression point.
The mechanical energy immediately following the collision is given by the equation:
E_f = .5mv_f^2 + .5kx^2
Substituting the given values:
E_f = .5(1 kg)(2.0 m/s)^2 + .5(200 N/m)(0.141 m)^2
E_f = 2 kg*m^2/s^2 + 1.9874 kg*m^2/s^2
E_f = 3.9874 kg*m^2/s^2
On the other hand, the mechanical energy at the maximum compression point is given by:
E_max = .5kx^2
Substituting the values:
E_max = .5(200 N/m)(0.141 m)^2
E_max = 1.9874 kg*m^2/s^2
Therefore, we can conclude that the mechanical energy immediately following the collision is equal to the mechanical energy at the maximum compression point of the block.
Now, let's move on to part e) and calculate x, the distance the block slides after the collision before turning around.
Using the work-energy principle, the work done by friction over that distance is equal to the change in kinetic energy of the block. The work done by friction is given by:
W = -umgx
Substituting the values:
W = -(0.10)(1.0 kg)(9.8 m/s^2) x
W = -0.98x
Since the work done is equal to the change in kinetic energy, we have:
W = ΔKE = KE_f - KE_i
The initial kinetic energy is zero since the block is initially at rest, and the final kinetic energy is given as:
KE_f = .5mv_f^2 = .5(1.0 kg)(2.0 m/s)^2
KE_f = 2 kg*m^2/s^2
Equating the work done by friction to the change in kinetic energy:
-0.98x = 2 kg*m^2/s^2
Solving for x:
x = (2 kg*m^2/s^2) / (-0.98)
x ≈ -2.04 m
However, distance cannot be negative in this context, so we can ignore the negative sign. Thus, the block slides approximately 2.04 meters after the collision before turning around.
Lastly, part f) asks for the maximum velocity reached by the block as it returns to its unstretched position following compression.
The maximum velocity can be found using the conservation of mechanical energy. At the maximum compression point, the mechanical energy is entirely in the form of potential energy stored in the spring:
E_max = .5kx^2
Substituting the values:
E_max = .5(200 N/m)(0.141 m)^2
E_max = 1.9874 kg*m^2/s^2
At the unstretched position, all of this potential energy is converted back into kinetic energy. Therefore:
KE_max = E_max
.5mv_max^2 = E_max
Substituting the mass and rearranging the equation:
.5(1 kg)v_max^2 = 1.9874 kg*m^2/s^2
v_max^2 = 3.9748 m^2/s^2
v_max = sqrt(3.9748) m/s
v_max ≈ 1.9937 m/s
Thus, the maximum velocity reached by the block as it returns to its unstretched position following compression is approximately 1.9937 m/s.