a rock with mass of 5kg is carried up a small hill 10 meters high. how much work had to be done in carrying the rock up hill
1 answer:
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a) The impulse is 76.5 Ns
b) The average force is 546.4 N
c) The final speed is 31.5 m/s
Explanation:
a)
The impulse exerted on an object is defined as
where
F is the magnitude of the force exerted on the object
is the time interval during which the force is applied
If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.
Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are
and whose height is
Therefore, the area (and the impulse) is
b)
In this problem, the force applied is not constant. However, we can rewrite the impulse also as
where
is the average force exerted during the whole time
In this problem we have
J = 76.5 Ns is the impulse (calculated in part a)
is the time interval
Solving for the average force, we find
c)
According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:
where
m is the mass of the object
v is the final velocity
u is the initial velocity
In this problem, we have
J = 76.5 Ns
m = 3.0 kg is the mass
u = 6.0 m/s is the initial velocity
Solving for v, we find the final velocity (and speed):
Learn more about impulse and momentum:
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Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²