The two blocks A and B have weights WA = 153 lbs and WB = 36 lbs. Starting from rest, the block A attains a speed of 3 ft/s after traveling 4ft down the incline having an angle q = 39.00°. Determine the kinetic coefficient of friction between the incline and block A, mk .
The Correct Answer and Explanation is :
To determine the coefficient of kinetic friction (μk) between block A and the incline, we can apply the principles of energy conservation and the work-energy theorem.
Step-by-step Solution:
Given:
- Weight of block A: ( W_A = 153 \, \text{lbs} )
- Weight of block B: ( W_B = 36 \, \text{lbs} )
- Final speed of block A: ( v = 3 \, \text{ft/s} )
- Distance traveled by block A: ( d = 4 \, \text{ft} )
- Angle of incline: ( \theta = 39^\circ )
- Initial speed of block A: ( u = 0 \, \text{ft/s} ) (since the block starts from rest)
The kinetic energy gained by block A is given by:
[
K.E. = \frac{1}{2} m v^2
]
where ( m ) is the mass of block A, and ( v ) is the final velocity.
To relate the mass of block A to its weight, we use the formula:
[
m = \frac{W_A}{g}
]
where ( g = 32.2 \, \text{ft/s}^2 ) is the acceleration due to gravity.
Now, let’s find the net work done on the block, considering the forces acting on block A.
The forces acting on block A are:
- The component of gravity pulling the block down the incline:
[
F_g = W_A \sin(\theta)
] - The frictional force opposing the motion of the block:
[
F_{\text{friction}} = \mu_k N
]
where ( N = W_A \cos(\theta) ) is the normal force on block A, and ( \mu_k ) is the coefficient of kinetic friction.
The net work ( W_{\text{net}} ) done on block A is the sum of the work done by gravity and the work done by friction:
[
W_{\text{net}} = F_g d – F_{\text{friction}} d
]
Substituting the expressions for ( F_g ) and ( F_{\text{friction}} ):
[
W_{\text{net}} = (W_A \sin(\theta)) d – (\mu_k W_A \cos(\theta)) d
]
By the work-energy theorem, the net work is also equal to the change in kinetic energy:
[
W_{\text{net}} = \Delta K.E.
]
Now, equating the two expressions:
[
(W_A \sin(\theta) – \mu_k W_A \cos(\theta)) d = \frac{1}{2} m v^2
]
Substitute ( m = \frac{W_A}{g} ) into the equation:
[
(W_A \sin(\theta) – \mu_k W_A \cos(\theta)) d = \frac{1}{2} \frac{W_A}{g} v^2
]
Simplifying and solving for ( \mu_k ):
[
\mu_k = \frac{W_A \sin(\theta) d – \frac{1}{2} \frac{W_A}{g} v^2}{W_A \cos(\theta) d}
]
Substituting the given values:
- ( W_A = 153 \, \text{lbs} )
- ( v = 3 \, \text{ft/s} )
- ( d = 4 \, \text{ft} )
- ( \theta = 39^\circ )
- ( g = 32.2 \, \text{ft/s}^2 )
We can now calculate ( \mu_k ).
Let me plug in the values for you to find the coefficient of kinetic friction.
The coefficient of kinetic friction (μk) between the incline and block A is approximately 0.765.
Explanation:
To find the coefficient of kinetic friction, we used the work-energy theorem, which relates the net work done on the block to its change in kinetic energy. The net work involves the work done by both gravity (which is trying to pull the block down the incline) and friction (which opposes the motion).
We started by writing the equation for the kinetic energy gained by block A as it moves down the incline. Then, we expressed the net work as the sum of the work done by gravity and the frictional force, and used this to solve for the coefficient of kinetic friction.
The final result of approximately 0.765 indicates the ratio of the frictional force to the normal force on block A while it slides down the incline. This value is crucial for understanding the block’s motion, and this frictional force affects how quickly the block accelerates down the incline.