How much work is required to push a 2.0-kg object up a frictionless inclines plane whose length is 2.0m and whose height is 1.0m
1 answer:
Answer:
20 Joules.
Explanation:
As , the incline is frictionless , only work done to push a block up the incline is only due to the gravitation resisting the upward movement .
Also, gravitational work only depends upon the height upto which the block is taken .
W = mgh .
Here, mass m of the object = 2.0 kg
g = 10m/
h = 1.0 m
Thus, W = 2×10×1 = 20 Joules.
So, 20 Joules of work is required to push an object of 2.0 kg mass up the frictionless incline .
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Answer:
1.64xrad/s
Explanation:
Given:
satellite is in elliptical orbit with a period T= 7.20 x s
Mass of planet m= 7.00x kg
Satellite's angular speed ωa- = 4.987 10-5 rad/s
At aphelion, at radius Ra= 5.1 x 10^7 m
Assuming torque is zero for this system, therefore, the conservation of angular momentum can be defines as
Lp = La
Ip ωp = Ia ωa--->eq(1)
Where,
'a' represents aphelion and 'p' represents perihelion
ω represents angular velocity
and I represents the rotational inertia
Since, I= 1/2 mR²
Here R is the radius at aphelion / perihelion
m = mass of planet
eq(1)=> ωp= Ia ωa/ Ip
ωp= (1/2 m Ra² ωa) / 1/2 m Rp²
ωp= (Ra/Rp)² ωa --->eq(2)
In order to find Rp, we use : 2a= Rp + Ra
where a represents semimajor axis
with the help of Kepler's third law for elliptical orbits
a= ∛(GmT²/4π²)
a= ∛ 6.67x x 7.00x
x (7.20 x
)² / 4π²
a= 3.97 x m
2a= Rp + Ra
Rp= 2a-Ra
Rp= 2 x 3.97 x - 5.1x
Rp= 2.84 x m
Substituting all the required values in eq 2, we have
ωp= (Ra/Rp)² ωa
ωp= (5.1x /2.84 x
)² x 4.987 x
ωp= 3.224 x 4.987 x
ωp= 1.64xrad/s
Therefore, angular speed at perihelion is 1.64xrad/s
Solution: 85.11 m
Given:
acceleration of the car,
initial velocity,
final velocity,
we need to find the displacement (s) of the car.
we would use the equation of motion:
hence, the displacement done by the car is = 85.11 m