A horizontal spring with k = 110 N/m has one end attached to the wall. A 290 g block is pushed onto the free end, compressing the spring by 0.160 m . The block is then released, and the spring launches it outward.
Neglecting friction, what's its speed when it leaves the spring?
Repeat previous part if the coefficient of kinetic friction is 0.300.
spring PE=initialKE
1/2 kx^2=1/2 m v^2
v= x sqrt k/m
To find the speed of the block when it leaves the spring, we can use the principle of conservation of energy. The potential energy stored in the compressed spring is converted into kinetic energy as the block is launched.
First, let's calculate the potential energy stored in the spring when it is compressed by 0.160 m.
Potential Energy (PE) = (1/2) * k * x^2
where k is the spring constant and x is the displacement of the spring.
PE = (1/2) * (110 N/m) * (0.160 m)^2
PE = 1.792 J
Next, we can equate the potential energy stored in the spring to the kinetic energy of the block as it leaves the spring.
Kinetic Energy (KE) = (1/2) * m * v^2
where m is the mass of the block and v is the velocity of the block.
Since there is no friction, the kinetic energy is equal to the potential energy. Therefore:
KE = 1.792 J
Now, we can solve for the velocity:
1.792 J = (1/2) * (0.290 kg) * v^2
v^2 = (2 * 1.792 J) / (0.290 kg)
v^2 = 12.4 m^2/s^2
v = sqrt(12.4 m^2/s^2)
v ≈ 3.52 m/s
Therefore, the speed of the block when it leaves the spring (neglecting friction) is approximately 3.52 m/s.
Now, let's repeat the calculation considering the coefficient of kinetic friction, which is 0.300.
When friction is present, the work done by friction needs to be taken into account. The work done by friction is given by:
Work by Friction = coefficient of kinetic friction * normal force * distance
In this case, the normal force is equal to the weight of the block, which is given by:
Normal Force = mass * gravity
where gravity is the acceleration due to gravity (approximately 9.8 m/s^2).
Normal Force = (0.290 kg) * (9.8 m/s^2)
Normal Force ≈ 2.842 N
Now, let's calculate the work done by friction:
Work by Friction = (0.300) * (2.842 N) * (0.160 m)
Work by Friction ≈ 0.137 J
Since work done by friction is equivalent to the loss in mechanical energy, the remaining kinetic energy will be:
KE = Potential Energy - Work by Friction
KE ≈ 1.792 J - 0.137 J
KE ≈ 1.655 J
Now, we can solve for the velocity:
1.655 J = (1/2) * (0.290 kg) * v^2
v^2 = (2 * 1.655 J) / (0.290 kg)
v^2 = 11.4 m^2/s^2
v = sqrt(11.4 m^2/s^2)
v ≈ 3.38 m/s
Therefore, the speed of the block when it leaves the spring (taking into account the coefficient of kinetic friction) is approximately 3.38 m/s.