Two blocks at rest on a frictionless inclined surface with mass of 3kg on a side inclined at 42° to the horizontal and mass 4.8kg a side inclined at 30° to the horizonal. if the two bodies are attached together by an extensible string passed through a pulley.
A) Determine the acceleration and the direction the blocks would move
B) How far do the blocks travel in 0.95s after release?
To determine the acceleration and direction of movement of the blocks, we need to analyze the forces acting on them.
A) Acceleration and direction of movement:
1. Start by analyzing the forces acting on each individual block independently.
2. For the 3kg block on the 42° inclined surface:
a. Resolve the weight force into components parallel and perpendicular to the inclined surface.
b. The component of the weight force parallel to the incline will cause the block to accelerate down the incline.
c. The perpendicular component of the weight force cancels out the normal force, leaving no vertical acceleration.
d. The friction force is absent on a frictionless surface.
e. Write Newton's second law equation: ΣF = ma, where ΣF is the sum of the forces acting on the block, m is the mass, and a is the acceleration.
f. The forces acting on the 3kg block are its weight component parallel to the incline (mgh*sin(42°)) and tension T (in the same direction as the acceleration).
g. The equation becomes: m*gsin(42°) - T = ma (Equation 1).
3. Repeat the process for the 4.8kg block on the 30° inclined surface:
a. Resolve the weight force into components parallel and perpendicular to the incline.
b. The component of the weight force parallel to the incline will cause the block to accelerate down the incline.
c. The perpendicular component cancels out the normal force, leaving no vertical acceleration.
d. The friction force is absent on a frictionless surface.
e. Write Newton's second law equation: ΣF = ma.
f. The forces acting on the 4.8kg block are its weight component parallel to the incline (mgh*sin(30°)) and tension T (in the opposite direction to the acceleration).
g. The equation becomes: T - m*gsin(30°) = ma (Equation 2).
4. Since the blocks are connected by an extensible string passing through a pulley, we know that the tension in the string is the same for both blocks.
Therefore, we can substitute T from Equation 1 into Equation 2: m*gsin(42°) - m*gsin(30°) = (m + M)*a, where M represents the mass of 4.8kg.
Rearranging the equation: a = (m*gsin(42°) - m*gsin(30°)) / (m + M).
5. Plug in the values: m = 3kg, M = 4.8kg, g = 9.8 m/s².
a = (3 * 9.8 * sin(42°) - 3 * 9.8 * sin(30°)) / (3 + 4.8) ≈ 1.28 m/s².
Therefore, the acceleration of the blocks is approximately 1.28 m/s².
Since the acceleration is positive, it indicates that the blocks will move in the direction of the 3kg block (down the incline with the 42° angle).
B) To calculate how far the blocks travel in 0.95 seconds after release, we can use the equations of motion under constant acceleration.
1. The formula to calculate the displacement (distance traveled) of an object under constant acceleration is:
s = ut + (1/2)at², where s is the displacement, u is the initial velocity (which is zero in this case), a is the acceleration, and t is the time.
2. Plug in the values: u = 0 m/s, a ≈ 1.28 m/s², t = 0.95 s.
s = 0 + (1/2) * 1.28 * (0.95)² ≈ 0.571 m.
Therefore, the blocks will travel approximately 0.571 meters in 0.95 seconds after release.