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
Superconductor, conductor, semiconductor, insulator
The distance at which the man slips is 0.3 m
Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.
Given-
mass of man= 70 kg
frictional coefficient μ=0.02
mass of body thrown= m2 = 3kg
let s be the stopping distance
we know that frictional force = F= μN
=μMg= 0.02 x 70 x 10
=14 N
∴acceleration, a= 14/70 = 0.2 m/s²
now on applying conservation of linear momentum
pi=pf pi=0 (initially at rest)
0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s
v1= m2v2 /m1= 0.3 m/s
we know,
v²- u² = -2as
0- (0.3) ²= -2 x 0.2 x 5
s= 0.09/0.4 ≈ 0.3 m
Learn more about distance here-
brainly.com/question/15172156
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