A 75N force is applied to a 15kg box. If the box
1 answer:
Answer:
The coefficient of kinetic friction is 0.51.
Explanation:
Given that,
Applied force, F = 75 N
Mass of a box, m = 15 kg
The acceleration of the box is 3 m/s²
We need to find the coefficient of kinetic friction. It is given by the formula as follows :
Where,
, is the normal force
So, the coefficient of kinetic friction is 0.51.
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Answer:
-9.46 N
Explanation:
In order to get the value of the x component of the resultant force, we need to get the value of the x component of each force.
This value will be the projection of the force vector, on the x-axis.
For F₁, as it is directed at an angle of 55.0º above the negative x axis, we can find F₁ₓ just applying the definition of cosine of an angle, as follows:
cos θ =
In this case, x = F₁ₓ, and r = F₁
θ, measured from the positive x axis counterclockwise, is as follows:
θ= 180º-55º = 125º
⇒ F₁ₓ = F₁* cosθ = 9.2 N * cos 125º = -5.28 N
We can repeat the process for F₂, as follows:
For F₂, as it is directed at an angle of 53.3º below the negative x axis, we can find F₂ₓ just applying the definition of cosine of an angle, as follows:
cos θ =
In this case, x = F₂ₓ, and r = F₂
θ, measured from the positive x axis counterclockwise, is as follows:
θ= 180º + 53.3º = 233.3º
⇒ F₂ₓ = F₂* cosθ = 7.00 N * cos 233.3º = -4.18 N
The total component of both forces along the x axis, can be found just adding both components, as follows:
Fₓ = F₁ₓ + F₂ₓ = -5.28 N + -4.18 N = -9.46 N