The student in item 1 moves the box up a ramp inclined at 12 degrees with the horizontal. If the box starts from rest at the bottom of the ramp and is pulled at an angle of 25 degrees with respect to the incline and with the same 185 N force, what is the acceleration up the the ramp? Assume that the coefficient of kinetic friction is 0.27
Question
Answer:
Acceleration up the ramp = -2.745 m/[tex]s^{2}[/tex]
Explanation:
Given
box is pulled at angle Θ = [tex]25^{o}[/tex]
Force applied F = 185N
coefficient of friction, μ[tex]_{k}[/tex] = 0.27
mass of the box m = 35 kg
We know that,
acceleration due to gravity g = 9.8 m/[tex]s^{2}[/tex]
horizontal component [tex]F_{x}[/tex] = F cosΘ = 185 * cos[tex]25^{o}[/tex] =167.67
vertical component [tex]F_{y}[/tex] = F sinΘ =185*sin[tex]25^{o}[/tex] = 78.18
Another vertical component is due to gravity [tex]F_{g}[/tex] , force in given by
[tex]F_{g}[/tex] = mg
= 35 x 9.8
= 343.35 N
Normal force [tex]F_{n}[/tex] = [tex]F_{g}[/tex] – [tex]F_{y}[/tex]
= 343.35 – 78.18
= 265.17 N
Frictional force [tex]F_{k}[/tex] = [tex]F_{n}[/tex] * μ[tex]_{k}[/tex]
= 265.17 * 0.27
= 71.596 N
To find acceleration, we know that,
force = mass x acceleration
acceleration = [tex]\frac{force}{mass}[/tex]
Here force is the summation of frictional force and horizontal component of the applied force. These force act in opposite directions.
force = [tex]F_{k}[/tex] – [tex]F_{x}[/tex]
= 71.596 – 167.67
= -96.074
acceleration = [tex]\frac{-96.074}{35}[/tex]
= -2.745 m/[tex]s^{2}[/tex]