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3 years ago

A 1kg object loses 20 j of gpe as it falls how far does it fall

1 answer:

4 0

Gravitational potential energy (GPE) is energy within an object with respect to its height. The equation of gpe is:

GPE = mgh

where:
m = mass of object
g = gravitational constant, 9.81 m/s^2
h = height

For a given gpe of 20 J, the height can be calculated as follows:
*Remember to be consistent with units. Since Joule is derived from SI units, make sure to use SI units for the other variables

GPE = 20 J = 1 (9.81) (h)
h = 2.04 meters

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balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

$a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)$

Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

$v_{f} ^{2} -v_{o} ^{2} = 2* a* \Delta x (2)$

Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

$\Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)$

We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

$v_{f} = v_{o} + a*\Delta t (4)$

Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

$\Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s (5)$

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