A 2.50 kg block is released from rest at point A as shown below. The track is frictionless except for the portion between points B and C, which has a length of 1.20 m. The block tr... A 2.50 kg block is released from rest at point A as shown below. The track is frictionless except for the portion between points B and C, which has a length of 1.20 m. The block travels down the track, hits a spring of spring constant 500 N/m, and compresses the spring 0.200 m from its equilibrium position before coming to rest momentarily. In this question, consider the system of the block, Earth, spring, and the track. (a) List the energy conversions in the system as the block moves (i) from A to B, (ii) from B to C, (iii) from C, hits the spring and comes to rest momentarily. (b) Calculate the kinetic energy of the block at point C. (c) Determine the coefficient of kinetic friction between the block and the rough surface BC. Read more

Steps to Solve

1. Energy conversions from A to B

Between points A and B, the block slides down a frictionless track. Therefore, the potential energy of the block is converted into kinetic energy.

2. Energy conversions from B to C

Between points B and C, the track has friction. Therefore, the kinetic energy of the block is converted into thermal energy (due to friction)

3. Energy conversions from C to spring compression

From point C until the spring is compressed, kinetic energy is converted into spring potential energy, and also thermal energy from the friction.

4. Calculate kinetic energy at point C

First, calculate the potential energy at point A with respect to point B, which will be converted entirely into kinetic energy at point B: $PE_A = mgh = (2.50 , \text{kg})(9.8 , \text{m/s}^2)(1.00 , \text{m}) = 24.5 , \text{J}$.

Next, the energy at B is equal to the energy lost by friction plus the spring potential energy: $KE_B = E_{\text{friction}} + PE_{\text{spring}}$.

The frictional force is $F_f = \mu_k mg$, where $\mu_k$ is the coefficient of kinetic friction. The work done by friction is $W_f = F_f d = \mu_k mg d$, where $d$ is the distance BC. At this point, we don't know $KE_C$ yet, but we know that the block compresses the spring by 0.200 m, so $PE_{\text{spring}} = (1/2)kx^2$. In this case, the block comes to rest momentarily after compressing the spring, meaning that energy from C is all lost to either friction, or used to compress the spring: $KE_C= E_{\text{friction}} + PE_{\text{spring}}$ Where the $E_{\text{friction}}$ is the energy lost to friction after point C until the spring is fully compressed.

The length of section BC is 1.20 m. The spring compresses an additional 0.200 m. Then the $E_{\text{friction}}=\mu_k m g (0.200 , \text{m})$ is the amount of energy lost to friction from when the block first touches the spring until it comes to a momentary stop. $PE_{\text{spring}} = \frac{1}{2}kx^2 = \frac{1}{2}(500 , \text{N/m})(0.200 , \text{m})^2 = 10 , \text{J}$. We will use these results in part (c).

To find the kinetic energy at point C ($KE_C$), use the following formula where $d$ is the length of BC: $KE_C = KE_B - W_f = KE_B - \mu_k mg d$ However, we do not know $\mu_k$, so we can't calculate $KE_C$ yet.

Recognize the system: the entire energy at A is equal to the frictional loss on BC, the frictional loss when compressing the spring, and spring potential energy at max compression: $PE_A=E_{\text{friction BC}} + E_{\text{friction spring}} + PE_{\text{spring}}$. $PE_A = \mu_k m g (1.20 , \text{m}) + \mu_k m g (0.200 , \text{m}) + 10 , \text{J}$ $24.5 , \text{J} = \mu_k (2.50 , \text{kg}) (9.8 , \text{m/s}^2) (1.20 , \text{m}) + \mu_k (2.50 , \text{kg}) (9.8 , \text{m/s}^2) (0.200 , \text{m}) + 10 , \text{J}$. Then we can solve for $\mu_k$ in the next part and go back to solve for $KE_C$, using the value for $PE_A$.

5. Determine the coefficient of kinetic friction

From the previous step, we can determine what $\mu_k$ is using $24.5 , \text{J} = \mu_k (2.50 , \text{kg}) (9.8 , \text{m/s}^2) (1.20 , \text{m}) + \mu_k (2.50 , \text{kg}) (9.8 , \text{m/s}^2) (0.200 , \text{m}) + 10 , \text{J}$. $24.5 , \text{J} - 10 , \text{J} = \mu_k (2.50 , \text{kg}) (9.8 , \text{m/s}^2) (1.20 , \text{m} + 0.200 , \text{m})$ $14.5 , \text{J} = \mu_k (2.50 , \text{kg}) (9.8 , \text{m/s}^2) (1.40 , \text{m})$ $14.5 , \text{J} = \mu_k (34.3 , \text{kg m}^2/\text{s}^2)$ $\mu_k = \frac{14.5 , \text{J}}{34.3 , \text{N m}} = 0.423$

6. Calculate kinetic energy at point C (continued)

Now that we have $\mu_k$ we can solve for $KE_C$ $KE_C = KE_B - W_f = PE_A - \mu_k mg d$ $KE_C = 24.5 , \text{J} - (0.423)(2.50 , \text{kg})(9.8 , \text{m/s}^2)(1.20 , \text{m})$ $KE_C = 24.5 , \text{J} - 12.4 , \text{J} = 12.1 , \text{J}$