I need a walk through please
1 answer:
Answer:
11,100 N
Explanation:
Draw the car on the incline. Label the weight and normal forces. Call the direction normal to the incline y, and the direction parallel to the incline x.
If it helps, rotate the incline so that the surface of the incline is flat.
Use trigonometry to determine the components of the weight force. The y component of the weight is:
Wᵧ = mg cos θ
Wᵧ = (1150 kg) (9.8 m/s²) (cos 8.70°)
Wᵧ = 11,100 N
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Answer:
2 m/s².
Explanation:
The forces acting on the crate have been shown in the figure below.
Net force acting on the crate in the horizontal direction = 400 - 250 = 150 N towards right.
For the forces of vertical direction we can write,
Normal force = 150 + Mg = 150 + 250 = 400 N (here g = 10 m/s² have been taken)
The magnitude of frictional force that will act on the crate = μ × Normal force
So, Frictional force = 0.25 × 400 = 100 N
Thus, the net force that will cause acceleration in the crate will be
Force in the right direction - frictional force
= 150 - 100 = 50 N
As we know,
Acceleration = =
= 2 m/s²
So, the net acceleration in the crate will be 2 m/s².
