Suppose there was 2 Newtons what would the mass of the cart be?
1 answer:
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a) The speed of the block at a height of 0.25 m is 2.38 m/s
b) The compression of the spring is 0.25 m
c) The final height of the block is 0.54 m
Explanation:
a)
We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:
where
is the initial potential energy, at the top
is the initial kinetic energy, at the top
is the final potential energy, at halfway
is the final kinetic energy, at halfway
The equation can be rewritten as
where:
m = 2.7 kg is the mass of the block
is the acceleration of gravity
is the initial height
u = 0 is the initial speed
is the final height of the block
v is the final speed when the block is at a height of 0.25 m
Solving for v,
b)
The total mechanical energy of the block can be calculated from the initial conditions, and it is
At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write
where
k = 453 N/m is the spring constant
x is the compression of the spring
And solving for x, we find
c)
If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:
while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:
where is the maximum height reached. Solving for this quantity, we find
which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.
Learn more about kinetic and potential energy:
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Answer:
A. The work done during the process is W = 0
B. The value of heat transfer during the process Q = 442.83
Explanation:
Given Data
Initial pressure = 100 k pa
Initial temperature = 25 degree Celsius = 298 Kelvin
Final pressure = 300 k pa
Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.
⇒ =
------------- (1)
Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.
⇒ P ∝ T
⇒ =
⇒ Put all the values in the above formula we get the final temperature
⇒ =
× 298
⇒ = 894 Kelvin
(A). Work done during the process is given by W = P × ()
From equation (1), =
so work done W = P × 0 = 0
⇒ W = 0
Therefore the work done during the process is zero.
Heat transfer during the process is given by the formula Q = m (
-
)
Where m = mass of the gas = 1 kg
= specific heat at constant volume of nitrogen = 0.743
Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )
⇒ Q = 442.83
Therefore the value of heat transfer during the process Q = 442.83
Refer to the figure shown below.
The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.
The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph
The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.
Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.
The velocity of the child and the velocity of the ship should be added vectorially to find the speed and direction of the child relative to the water surface.
The magnitude of the child's velocity is
v = √(2² + 18²) = 18.11 mph
The direction of the child's speed is
θ = tan⁻¹ (18/2) = tan⁻¹ 9 = 83.7° north of east or counterclockwise from the eastern direction.
Answer:
The magnitude is 18.1 mph.
The direction is 84° north of east.