Respuesta :
(a) Work done on the oven = 110J
(b) Work done by the friction force =250J
(c) Increase in PE =1000J
(d) Increase in KE = 640J
(e) Acceleration of the oven =14.9 m/s² And KE =74.5J
Given
g = 10
mass = 10kg
distance = 8 m
Magnitude of force = 110 N
Kinetic friction = 0.250 N
A
The work done on the oven by the force is given by
W = F.d
W = 110*14cos(0)
It is a Horizontal component
W = 110 J
B
Now to calculate the work done by the friction force
Frictional force = μ * N
Frictional force = 0.25*10*10
Frictional force = 25 N
Frictional work = frictional force . d
work(f) = 25 * 10cos(0)
Work(f) = 250 J
C
Increase in the Potential energy
PE = mgh
PE = 10*10*10cos(0)
PE = 1000 J
D
The increase in the oven's kinetic energy
PE₁ = KE₁ + W = PE₂ + KE₂
110 - 260 - 1000 = KE₂
KE₂ = 640J
E
The acceleration of the oven
Force - frictional force = ma
a = 110 -250/10 = 14 m/s²
Now to calculate KE we should know the velocity
V(f)² = v(i)² + 2ad
v(f)² = 2(14)(8) = 224
v(f) = 14.9 m/s
Now KE will be given as
(1/2)mv²
KE = (1/2)(10)(14.9)
KE = 74.5 J
Therefore, we got to know the work done, PE and KE just by the force applied.
To know more about the force, refer: https://brainly.com/question/13606948
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