A cookie jar is moving up a `40^(@)` incline. At a point 45 cm from the bottom of the incline ( measured along the incline ) ,the jar has a speed of 1.4 m/s. The coefficient of kinetic friction between jar and incline is 0.15 . (a) How much farther up the incline will the jar move ? (b) How fast will it be going when it has slid back to the bottom of the incline ? (c) Do the answers to parts (a) and (b) increase, decrease, or remain the same if we decrease the coefficient of kinetic friction ( but do not change the given speed or location ) ?

Answer

To solve the problem step by step, we will break it down into parts (a), (b), and (c) as outlined in the question.

### Part (a): How much farther up the incline will the jar move?

1. **Identify the Forces Acting on the Jar:**
- The gravitational force acting down the incline: \( F_g = mg \sin \theta \)
- The frictional force acting down the incline: \( F_f = \mu mg \cos \theta \)
- The total deceleration \( a \) when the jar is moving up is given by:
\[ a = g \sin \theta + \mu g \cos \theta \]

2. **Calculate the Acceleration:**
- Given \( \theta = 40^\circ \), \( \mu = 0.15 \), and \( g = 9.81 \, \text{m/s}^2 \):
\[ a = g \sin(40^\circ) + \mu g \cos(40^\circ) \]
- Calculate \( \sin(40^\circ) \) and \( \cos(40^\circ) \):
\[ \sin(40^\circ) \approx 0.6428, \quad \cos(40^\circ) \approx 0.7660 \]
- Substitute these values into the acceleration equation:
\[ a = 9.81 \times 0.6428 + 0.15 \times 9.81 \times 0.7660 \]
\[ a \approx 6.303 + 1.125 \approx 7.428 \, \text{m/s}^2 \]

3. **Use the Kinematic Equation:**
- The initial speed \( u = 1.4 \, \text{m/s} \) and final speed \( v = 0 \, \text{m/s} \) when the jar stops.
- Using the equation \( v^2 = u^2 + 2as \):
\[ 0 = (1.4)^2 - 2(7.428)s \]
\[ 0 = 1.96 - 14.856s \]
\[ 14.856s = 1.96 \quad \Rightarrow \quad s = \frac{1.96}{14.856} \approx 0.132 \, \text{m} \approx 13.2 \, \text{cm} \]

4. **Total Distance Up the Incline:**
- The jar moves an additional \( 13.2 \, \text{cm} \) up the incline.

### Part (b): How fast will it be going when it has slid back to the bottom of the incline?

1. **Calculate the Total Distance Down:**
- The total distance down is the distance it initially moved up (45 cm) plus the distance it moved up before stopping (13.2 cm):
\[ \text{Total distance down} = 0.45 \, \text{m} + 0.132 \, \text{m} = 0.582 \, \text{m} \]

2. **Calculate the Acceleration Going Down:**
- When the jar slides down, the acceleration is:
\[ a' = g \sin \theta - \mu g \cos \theta \]
- Substitute the values:
\[ a' = 9.81 \times 0.6428 - 0.15 \times 9.81 \times 0.7660 \]
\[ a' \approx 6.303 - 1.125 \approx 5.178 \, \text{m/s}^2 \]

3. **Use the Kinematic Equation Again:**
- Initial speed \( u = 0 \, \text{m/s} \) at the top, final speed \( v \) at the bottom:
\[ v^2 = u^2 + 2a's \]
\[ v^2 = 0 + 2(5.178)(0.582) \]
\[ v^2 = 6.048 \quad \Rightarrow \quad v = \sqrt{6.048} \approx 2.46 \, \text{m/s} \]

### Part (c): Effect of Decreasing the Coefficient of Kinetic Friction

1. **Distance Up the Incline:**
- If the coefficient of kinetic friction \( \mu \) decreases, the deceleration \( a \) will decrease, leading to a greater distance traveled up the incline.

2. **Velocity at the Bottom:**
- Similarly, if \( \mu \) decreases, the acceleration \( a' \) when sliding down will increase, leading to a higher final velocity when the jar reaches the bottom.

### Summary of Answers:
- (a) The jar will move approximately **13.2 cm** farther up the incline.
- (b) The speed when it slides back to the bottom will be approximately **2.46 m/s**.
- (c) Both the distance traveled up the incline and the final velocity at the bottom will **increase** if the coefficient of kinetic friction is decreased.