A 15-kg block is on a frictionless ramp that is inclined at 20° above the horizontal. It is connected by a very light string over an ideal pulley at the top edge of the ramp to a hanging 19-kg block, as shown in the figure. The string pulls on the 15-kg block parallel to the surface of the ramp. Find the magnitude of the acceleration (in m/s2) of the 19-kg block after the system is gently released?
4.0m/s^2
To find the magnitude of the acceleration of the 19-kg block, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's break down the problem step by step:
1. Start by drawing a free-body diagram for each block.
For the 15-kg block:
- There is a tension force acting to the right, parallel to the ramp.
- The weight of the block acts vertically downward.
For the 19-kg block:
- There is a tension force acting vertically upward.
- The weight of the block acts vertically downward.
2. Determine the forces acting on each block.
For the 15-kg block:
- The weight of the block (mg) has two components:
- The component parallel to the ramp is mg * sin(theta), where theta is 20 degrees.
- The component perpendicular to the ramp is mg * cos(theta).
- The tension force acts parallel to the ramp.
For the 19-kg block:
- The weight of the block is mg, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
- The tension force acts vertically upward.
3. Apply Newton's second law to each block.
For the 15-kg block:
- The net force in the horizontal direction is the tension force (T) acting to the right minus the component of the weight force parallel to the ramp, which is mg * sin(theta).
- Applying Newton's second law, we have:
T - mg * sin(theta) = ma, where a is the acceleration of the system.
For the 19-kg block:
- The net force in the vertical direction is the tension force acting upward minus the weight force acting downward, which is mg.
- Applying Newton's second law, we have:
T - mg = ma.
4. Solve the system of equations.
Now we have two equations with two unknowns, T (tension force) and a (acceleration). We can solve these equations simultaneously to find the desired acceleration of the 19-kg block.
Equation 1: T - mg * sin(theta) = ma
Equation 2: T - mg = ma
Rearrange equation 1 to solve for T:
T = mg * sin(theta) + ma
Substitute T into equation 2:
mg * sin(theta) + ma - mg = ma
mg * sin(theta) = mg
g * sin(theta) = g
sin(theta) = 1
Since sin(20 degrees) = 1/2, we can simplify the equation to:
g/2 = g
1/2 = 1
This means that the magnitude of the acceleration of the 19-kg block is equal to the acceleration due to gravity, which is approximately 9.8 m/s^2.
Therefore, the magnitude of the acceleration of the 19-kg block after the system is gently released is 9.8 m/s^2.
To find the magnitude of acceleration of the 19-kg block, we need to consider the forces acting on both blocks.
Let's start by analyzing the forces on the 15-kg block. Since the ramp is frictionless, the only forces acting on the 15-kg block are its weight (mg) pointing straight down and the tension force (T) from the string pulling it parallel to the ramp.
The weight force can be resolved into two components: one perpendicular to the ramp (mg * cosθ) and one parallel to the ramp (mg * sinθ). Here, θ is the angle of inclination of the ramp (20°).
Now let's analyze the forces acting on the 19-kg block. Again, its weight (19g) points straight down, and there is tension force (T) in the string pulling it upwards.
Since the blocks are connected by a string, the tension force (T) acting on both blocks is the same.
Now, let's write down the equations of motion for each block using Newton's second law:
For the 15-kg block:
Sum of forces = (mass * acceleration)
T - mg * sinθ = 15a (equation 1)
For the 19-kg block:
Sum of forces = (mass * acceleration)
19g - T = 19a (equation 2)
Solving these two equations simultaneously will give us the value of acceleration (a).
First, let's eliminate T from the equations by adding equations 1 and 2:
(T - mg * sinθ) + (19g - T) = 15a + 19a
Simplifying the equation:
19g - mg * sinθ = 34a
Now, substitute the values:
(19 * 9.8) - (15 * 9.8 * sin(20°)) = 34a
Calculating the values:
186.2 - (15 * 9.8 * 0.342) = 34a
186.2 - 50.43 = 34a
135.77 = 34a
Finally, solve for a:
a = 135.77 / 34
Calculating the value of a:
a ≈ 3.99 m/s^2
Therefore, the magnitude of acceleration of the 19-kg block is approximately 3.99 m/s^2.