A block of mass m = 5.9 kg is pulled up a è = 21° incline as in the figure with a force of magnitude F = 34 N.
(a) Find the acceleration of the block if the incline is frictionless.
1
Your response differs from the correct answer by more than 10%. Double check your calculations. m/s2
(b) Find the acceleration of the block if the coefficient of kinetic friction between the block and incline is 0.12.
Wb = mg = 5.9kg * 9.8N/kg = 57.82N. =
Weight of block.
Fb = (57.82N,21deg.) = Force of block.
Fp = 57.82sin21 = 20.72N. = Force parallel to the incline.
Fv = 57.82cos21 = 53.98N. = Force perpendicular to incline. = Tne Normal.
Fn = Fap-Fp = 34 - 20.72 = 13.28N. =
Net force. Fap = Force applied.
a. Fn = ma,
a = Fn/m = 13.28 / 5.9 = 2.25m/s^2.
b. Ff = u*Fv = 0.12 * 53.98 = 6.48N. =
Force of friction.
Fn = Fap - Fp -Ff = 34 - 20.72 - 6.48 =
6.8N.
a = 6.8 / 5.9 = 1.15m/s^2.
(a) To find the acceleration of the block when the incline is frictionless, we will use the formula:
\[ \sin(\theta) = \frac{{\text{{opposite}}}}{{\text{{hypotenuse}}}} \]
In this case, the force of magnitude F is acting as the opposite side of the right triangle, and the weight of the block (mg) is acting as the hypotenuse.
Plugging in the values, we have:
\[ \sin(21°) = \frac{{F}}{{mg}} \]
Solving for mg:
\[ mg = \frac{{F}}{{\sin(21°)}} \]
Substituting the given values:
\[ mg = \frac{{34 N}}{{\sin(21°)}} \]
Now, we can solve for the acceleration using Newton's second law:
\[ F_{\text{{net}}} = ma \]
Since the block is being pulled up the incline, the net force is equal to F (the force pulling the block). So,
\[ F_{\text{{net}}} = F = ma \]
Substituting the given values:
\[ 34 N = (5.9 kg) \cdot a \]
Solving for a:
\[ a = \frac{{34 N}}{{5.9 kg}} \]
Calculating this value, we get:
\[ a \approx 5.76 \, \text{{m/s}}^2 \]
Therefore, the acceleration of the block is approximately 5.76 m/s^2.
(b) When the coefficient of kinetic friction between the block and incline is 0.12, we need to consider the additional force of friction.
The force of friction can be calculated using the formula:
\[ F_{\text{{friction}}} = \mu \cdot N \]
where N is the normal force and is equal to mg, and μ is the coefficient of kinetic friction.
Plugging in the given values, we have:
\[ F_{\text{{friction}}} = (0.12) \cdot mg \]
Substituting the value of mg, we get:
\[ F_{\text{{friction}}} = (0.12) \cdot (5.9 kg) \cdot g \]
where g is the acceleration due to gravity.
The net force acting on the block will now be:
\[ F_{\text{{net}}} = F - F_{\text{{friction}}} \]
Substituting the given values:
\[ F_{\text{{net}}} = 34 N - (0.12) \cdot (5.9 kg) \cdot g \]
Again, using Newton's second law:
\[ F_{\text{{net}}} = ma \]
Substituting the value of F_{\text{{net}}}:
\[ 34 N - (0.12) \cdot (5.9 kg) \cdot g = (5.9 kg) \cdot a \]
Now, we can solve for the acceleration of the block:
\[ a = \frac{{34 N - (0.12) \cdot (5.9 kg) \cdot g}}{{5.9 kg}} \]
Substituting the value of g and calculating this expression, we get:
\[ a \approx 3.05 \, \text{{m/s}}^2 \]
Therefore, the acceleration of the block when the coefficient of kinetic friction is 0.12 is approximately 3.05 m/s^2.
To find the acceleration of the block in both scenarios, we can use Newton's second law of motion. This law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
(a) If the incline is frictionless, the only force acting on the block is the force applied in the upward direction. The gravitational force acting on the block is perpendicular to the direction of motion and does not affect it. Therefore, the net force is equal to the applied force.
Net force = Applied force
m * a = F
Substituting the given values:
m = 5.9 kg
F = 34 N
5.9 kg * a = 34 N
To find the acceleration, divide both sides of the equation by the mass:
a = 34 N / 5.9 kg
Calculating a, we get:
a ≈ 5.76 m/s² (rounded to two decimal places)
(b) If the coefficient of kinetic friction between the block and the incline is 0.12, we need to consider the frictional force opposing the motion.
The force of friction on an inclined plane can be calculated as:
Frictional force = Coefficient of friction * Normal force
The normal force is the component of the weight of the object perpendicular to the incline and can be calculated as:
Normal force = m * g * cos(θ)
Given:
m = 5.9 kg
g = 9.8 m/s² (acceleration due to gravity)
θ = 21° (angle of the incline)
Coefficient of kinetic friction = 0.12
First, calculate the normal force:
Normal force = 5.9 kg * 9.8 m/s² * cos(21°)
Then, calculate the frictional force:
Frictional force = 0.12 * Normal force
Now, the net force can be obtained by subtracting the frictional force from the force applied:
Net force = Applied force - Frictional force
Finally, use Newton's second law to find the acceleration:
m * a = Net force
Solve for a to find the acceleration.