Answer:
630.75 j
Explanation:
from the question we have the following
total mass (m) = 54.5 kg
initial speed (Vi) = 1.4 m/s
final speed (Vf) = 6.6 m/s
frictional force (FF) = 41 N
height of slope (h) = 2.1 m
length of slope (d) = 12.4 m
acceleration due to gravity (g) = 9.8 m/s^2
work done (wd) = ?
- we can calculate the work done by the boy in pushing the chair using the law of law of conservation of energy
wd + mgh = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d)
wd = (0.5 mVf^2) - (0.5 mVi^2) + (FF x d) - (mgh)
where wd = work done
m = mass
h = height
g = acceleration due to gravity
FF = frictional force
d = distance
Vf and Vi = final and initial velocity
wd = (0.5 x 54.5 x 6.9^2) - (0.5 x 54.5 x 1.4^2) + (41 x 12.4) - (54.5 X 9.8 X 2.1)
wd = 630.75 j
Answer:T=1316.21 N
Explanation:
The tension has two components: Vertical and Horizontal. The
horizontal component is ma, the vertical component is mg. Using
Pythagoras theorem, we can find the tension as:
T=((ma)^2 (mg)^2)^(1/2)
So
T=((129*2.84)^2 (129*9.8)^2)^(1/2)
T=1316.21 N
Momentum = mass x velocity
12 = 4 x v | ÷ both sides by 4
12 ÷ 4 =v
v= 3 m/s
Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN
