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

Since the track is frictionless between A and B, the block's gravitational potential energy is converted entirely into kinetic energy.

2. Energy conversions from B to C

As the block moves from B to C, kinetic energy is converted into thermal energy due to the friction between the block and the surface.

3. Energy conversions from C to spring compression

From C to the point where the spring is maximally compressed, the block's kinetic energy is converted into spring potential energy and thermal energy due to friction.

4. Kinetic energy at point B

First we can calulate the potential energy at point A $PE = mgh = 2.50 \text{ kg} \cdot 9.8 \text{ m/s}^2 \cdot 1.00 \text{ m} = 24.5 \text{ J}$

Since the track is frictionless from A to B, all potential energy at point A will be converted to kinetic energy at point B. Therefore, kinetic energy at point B is $24.5 \text{ J}$.

5. Work done by friction

Final Spring Potential Energy: $$SPE = \frac{1}{2}kx^2 = \frac{1}{2} \cdot 500 \text{ N/m} \cdot (0.200 \text{ m})^2 = 10 \text{ J}$$

The block comes to rest, so the final kinetic energy is zero. The total energy at B is equal to the elastic potential energy in the spring at the end, plus the work done by friction $W_f$ along BC: $KE_B = SPE + W_f$, where $KE_B$ is kinetic energy at point B, $SPE$ is spring potential energy, and $W_f$ is work done by friction

Therefore $24.5 \text{ J} = 10 \text{ J} + W_f$ $W_f = 14.5 \text{ J}$

6. Kinetic energy at point C

The work done by friction, $W_f = KE_B - KE_C$, therefore $KE_C = KE_B - W_f$

The work done by friction can also be expressed as: $W_f = F_f \cdot d = KE_B - KE_C$, where $d$ is the distance from B to C.

From step 5 we know that $W_f = 14.5 \text{ J}$ between B and C, and when compressing the spring. The work done by friction only between B and C will exclude the compression of the spring: $W_{f_{BC}} = KE_B - SPE - KE_C$

Since the block is momentarily at rest, the kinetic energy before it hits the spring will be converted into spring potential energy, and the work done by friction along BC: $KE_C = SPE + W_{f_{spring}}$

$W_{f_{total}} = W_{f_{BC}} + W_{f_{spring}}$ where $W_{f_{total}} = 14.5 \text{ J}$, and $SPE = 10 \text{ J}$

$W_{f_{BC}} = 14.5 \text{ J} - W_{f_{spring}}$ but $W_{f_{spring}}$ can be rewritten as $W_{f_{spring}} = KE_C - SPE$

So substituting $W_{f_{spring}}$ with $KE_C - SPE$ gives us $W_{f_{BC}} = 14.5 \text{ J} - (KE_C - SPE)$ From Step 5, rearrange $KE_B = SPE + W_{f_{BC}}$, to solve for $KE_C$ at point C: $KE_C = KE_B - W_{f_{BC}} = 24.5 \text{ J} - W_{f_{BC}}$ $KE_C = 24.5 \text{ J} - W_{f_{BC}}$, where $W_{f_{BC}}$ is the work done by friction between points $B$ and $C$. To find $KE_C$, we first need to calculate $W_{f_{BC}} = KE_C + SPE = 14.5 \text{ J}$

Since all kinetic energy will be converted to spring potential energy and also work done by friction when compressing the spring, we can express the equation as: $KE_C - W_{f_{spring}} = SPE$ becomes $KE_C = SPE + W_{f_{spring}}$ $KE_C = 10 \text{ J} + W_{f_{spring}}$ so with this equation we can solve for $KE_C$ by finding $W_{f_{spring}}$ To solve for $W_{f_{spring}}$, we can rearrange the equation

The remaining term given by $KE_B - KE_C = W_{f_{BC}}=24.5 \text{ J} - KE_C$

$KE_C = 24.5 \text{ J} - W_{f_{BC}}$ we can solve for $KE_C$ if $W_{f_{BC}}$ can be found. But $W_{f_{BC}}$ can be found by $W_{f_{BC}} = F_f \cdot d$ so the force of friction over segment BC is $$ F_f = \mu_k \cdot N = \mu_k \cdot mg $$

So therefore: $W_{f_{BC}} = \mu_k \cdot mg \cdot d$. So if we find $\mu_k$, we can solve

Since $KE_B = KE_C + W_f$ therefore: $KE_C = KE_B - W_f$ where $W_f = \mu * m * g * d$, where d is 1.2 m Since $KE_B = 24.5 text{ J}$ $$24.5 = KE_C + \mu * m * g * d$$ Since $SPE = KE_C - W_f = text{ J}$, where frictional work $W_f$ does not include the rough serface of BC However, $SPE=1/2 * k *x^2 = 0.5 * 500 * 0.2^2 = 10 text{ J}$, so

From Step 5, $KE_C = 24.5 \text{ J} - W_{f_{BC}}$. To solve, $KE_C$ must first be calculated:

$KE_B + W_f = SPE$, therefore $KE_B = SPE - W_f$. But spring force cannot be negative, thus

$KE_B = KE_C + F_f * d = KE_C + 14.5$, therefore $KE_C = KE_B - F_f * d = 24.5 - 14.5$, and therefore $KE_C = 10 text{ J}$. Final KE at point C = 10 J.

7. Coefficient of kinetic friction

The work done by friction between B and C can be calculated as $W_f = F_f \cdot d = \mu_k \cdot m \cdot g \cdot d$, where $\mu_k$ is the coefficient of kinetic friction, $m$ is the mass, $g$ is the acceleration due to gravity, and $d$ is the length of BC. At point C, the block has kinetic energy. During the compression of the spring, the kinetic energy at C is dissipated by the work done on the spring, and the friction. therefore, $KE_C$ will equal the $W_f$

$KE_C = 10 \text{ J}$, $d = 1.20 \text{ m}$, which we can use to find the coefficient.

$W_f = KE_B - SPE_{\text{spring}}$, where $SPE = 1/2 kx^2 = 1/2 (500)(.2)^2 = 10 text{ J}$ Then $KE_C = KE_B - u_k m g d$ that can be rearranged to $24.5 \text{ J} - u_k (2.5)(9.8)(1.2)$, so $u_k = 10/((2.5)(9.8)(1.2))$ which simplifies to The kinetic energy at C is used to compress the spring and overcome the friction Spring Potential energy is $1/2kx^2=1/2500.2^2=10J$

$$W_{BC} = KE_i - KE_f$$ $KE_B=24.5J$ and $KE_C$ will be found: Final step is $24.5J -KE_C=(\mu)(2.5kg)(9.8 text{ m/s}^2)(1.2m)$ with $KE_C=10 \text{ J}$ then $\mu=0.49$