A man exerts a constant force to pull a 51-kg box across a floor at constant speed. He exerts this force by attaching a rope to
the box and pulling so that the rope makes a constant angle of 36.9∘ above the horizontal. The coefficient of kinetic friction for the box-floor interface is μk = 0.12. What is the work done by the man if he moves the box 10m
1 answer:
Answer:
W=561.41 J
Explanation:
Given that
m = 51 kg
μk = 0.12
θ = 36.9∘
Lets F is the force applied by man
Given that block is moving at constant speed it mans that acceleration is zero.
Horizontal force = F cos θ
Vertical force = F sinθ
Friction force Fr= μk N
N + F sinθ = m g
N = m g - F sinθ
Fr = μk (m g - F sinθ)
For equilibrium
F cos θ = μk (m g - F sinθ)
F ( cos θ +μk sinθ) = μk (m g
Now by putting the values
F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10
F= 70.2 N
We know that Work
W= F cos θ .d
W= 70.2 x cos 36.9∘ x 10
W=561.41 J
You might be interested in
Answer:
Φ = 361872 N.m^2 / C
Explanation:
Given:-
- The area of the two plates,
- The charge on each plate,
- Permittivity of free space,
- The radius for the flux region,
- The angle between normal to region and perpendicular to plates, θ = 4°
Find:-
Find the flux (in N · m2/C) through a circle of radius 3.3 cm between the plates.
Solution:-
- First we will determine the area of the region ( Ar ) by using the formula for the area of a circle as follows. The region has a radius of r = 3.3 cm:
- The charge density ( σ ) would be considered to be uniform for both plates. It is expressed as the ratio of the charge ( q ) on each plate and its area ( A_p ):
σ =
σ = 0.00094 C / m^2
- We will assume the electric field due to the positive charged plate ( E+ ) / negative charged plate ( E- ) to be equivalent to the electric field ( E ) of an infinitely large charged plate with uniform charge density.
- The electric field experienced by a region between two infinitely long charged plates with uniform charge density is the resultant effect of both plates. So from the principle of super-position we have the following net uniform electric field ( E_net ) between the two plates:
- From the Gauss-Law the flux ( Φ ) through a region under uniform electric field ( E_net ) at an angle of ( θ ) is:
Φ = E_net * Ar * cos ( θ )
Φ = (106214689.26553) * (0.00342) * cos ( 5 )
Φ = 361872 N.m^2 / C