We need to see what forces act on the box:
In the x direction:
Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.
In the y direction:
N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.
From N-Gcosα=0 we get:
N=Gcosα, we will need this for the force of friction.
Now to solve for Fh:
Fh=ma + Ff + Gsinα,
Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²
Fh=ma + μmgcosα+mgsinα
Now we plug in the numbers and get:
Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N
The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
Answer:
1-D(carbon dioxide, water and sunlight)
2-D(parasitism)
3-C(competition)
Explanation:
hope it helps
Answer:
v = 8.45 m/s
Explanation:
given,
mass = 3 kg
angle = 30.0°
vertical distance = 3.3 m
μ = 0.06
according to conservation of energy
KE(loss) = PE(gain) + Work done (against\ friction)..............(1)
frictional Force
work against friction
W = F d
Potential energy
PE = mgh
v = 8.45 m/s
the minimum speed is equal to 8.45 m/s
