A 15 kg block has a constant acceleration of [tex]2.2 \, \text{m/s}^2[/tex] down a 30-degree incline.
What is the magnitude of the friction force on the block?
Answer (3)
Use Newton's Second law of motion and plug in what you know: 15sin30-friction=15a 15 times 9.8 times 0.5-15 times 2.2 =Friction So Friction = 73.5 -33, or 40.5 Newtons Hope this helps!
A 15kg block has a** constant acceleration** of 2.2m/s² down a 30-degree incline, The magnitude of the friction force on the **block **is 40.11 N.
The component of the force of gravity (weight) parallel to the incline:
F(parallel) = m × g × sin(θ)
Where:
m = **mass **of the block (15 kg)
g = acceleration due to** gravity** (approximately 9.81 m/s²)
θ = angle of the incline (30 degrees)
F(parallel) = 15 × 9.81 × sin(30°)
F(parallel) = 73.11 N
The net force along the incline:
Net force = m × a
Net force = 15 × 2.2
Net force = 33 N
The** friction force** (opposing the motion):
Friction force = Net force - F(parallel0
Friction force = 33 - 73.11
**Friction force **= -40.11 N
Therefore, The magnitude of the** friction force **on the block is 40.11 N.
To know more about the friction force:
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The friction force acting on the block is approximately 40.58 N. This is calculated by resolving the gravitational force into components and applying Newton's second law. The net force down the incline is determined, and the friction force opposing this motion is found accordingly.
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