A 1.70-kg block slides on a rough, horizontal surface. The block hits a spring with a speed of 2.00 m/s and compresses it a distance of 11.3 cm before coming to rest. If the coefficient of kinetic friction between the block and the surface is μk = 0.534, what is the force constant of the spring?
To find the force constant of the spring, we can use the concept of conservation of mechanical energy. Initially, the block has kinetic energy due to its initial velocity, and when it hits the spring, this kinetic energy is converted into potential energy stored in the compressed spring.
The formula for the potential energy stored in a spring is given by:
PE = (1/2)kx^2
where PE is the potential energy, k is the force constant of the spring, and x is the displacement or compression of the spring.
In this case, the block comes to rest after compressing the spring. Therefore, the potential energy stored in the spring is equal to the initial kinetic energy of the block.
Initial kinetic energy (KE) = PE
The initial kinetic energy is given by:
KE = (1/2)mv^2
where m is the mass of the block and v is the initial velocity of the block.
Substituting the given values:
m = 1.70 kg
v = 2.00 m/s
KE = (1/2)(1.70 kg)(2.00 m/s)^2
KE = 1.70 J
Now we can use this value of the initial kinetic energy to calculate the force constant of the spring.
PE = (1/2)kx^2
We know that the potential energy (PE) is equal to the initial kinetic energy (KE) because the block comes to rest.
1.70 J = (1/2)k(0.113 m)^2
1.70 J = (1/2)k(0.012769 m^2)
Now we can solve for k by rearranging the equation:
k = (2 * 1.70 J) / (0.012769 m^2)
k ≈ 266.13 N/m
Therefore, the force constant of the spring is approximately 266.13 N/m.