A block with a mass of m = 21 kg rests on a frictionless surface and is subject to two forces acting on it. The first force is directed in the negative x-direction with a magnitude of F1 = 9.5 N. The second has a magnitude of F2 = 19.5 N and acts on the body at an angle θ = 10 ° up from the horizontal as shown.
m = 21 kg
F1 = 9.5 N
F2 = 19.5 N
θ = 10 °
Solve numerically for the block's acceleration in the x-direction, ax, in m/s2.
ax=
F = F1 + F2 = -9.5 + 19.5[10o]
F = -9.5 + 19.5*cos10 + (19.5*sin10)
F = -9.5 + 19.20 + 3.39i
F = 9.7 + 3.39i
a = X/m = 9.7/21 = 0.462 m/s^2.
To solve for the block's acceleration in the x-direction (ax), we can use Newton's second law of motion.
First, let's break down the forces acting on the block:
1. The force F1 is directed in the negative x-direction, so its x-component is -F1.
2. The force F2 is at an angle θ = 10° up from the horizontal. To find its x-component, we can use the formula: F2(cosθ). Therefore, the x-component of F2 is F2 cosθ.
Next, we can calculate the net force in the x-direction:
Net force in the x-direction = -F1 + F2 cosθ
Since the block is subject to no friction, the net force is equal to the mass of the block multiplied by its acceleration:
Net force in the x-direction = m * ax
Equating the two expressions for the net force in the x-direction, we get:
m * ax = -F1 + F2 cosθ
Now, let's substitute the given values into the equation:
m = 21 kg
F1 = 9.5 N
F2 = 19.5 N
θ = 10 °
Plugging in the values, we have:
21 kg * ax = -9.5 N + 19.5 N * cos(10°)
Using a calculator to calculate cos(10°), we get:
21 kg * ax = -9.5 N + 19.5 N * 0.9848
21 kg * ax ≈ -9.5 N + 19.1856 N
21 kg * ax ≈ 9.6856 N
Dividing both sides of the equation by 21 kg, we find:
ax ≈ 0.46122 m/s²
Therefore, the block's acceleration in the x-direction is approximately 0.46122 m/s².
To solve for the block's acceleration in the x-direction, we can use Newton's second law of motion, which states that the net force on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it can be written as:
ΣF = m * a
In this case, there are two forces acting on the block: F1 in the negative x-direction and F2 at an angle θ. We need to break down the force F2 into its x and y components.
The x-component of F2 can be calculated using the equation:
F2x = F2 * cos(θ)
Substituting the given values:
F2x = 19.5 N * cos(10°)
Next, we can determine the net force in the x-direction by considering the direction of each force. Since F1 is in the negative x-direction and F2x is in the positive x-direction, the net force in the x-direction can be calculated as:
ΣFx = F1 - F2x
Substituting the given values:
ΣFx = 9.5 N - (19.5 N * cos(10°))
Finally, to solve for the acceleration in the x-direction (ax), we can rearrange Newton's second law equation as:
ax = ΣFx / m
Substituting the known values, we can now solve for ax numerically.