How far does a 1.58-kg stone with the kinetic energy of 3.11 J go in 1.86 s if it is moving in a straight line?
To determine how far the stone will go, we need to use the equation for kinetic energy and the kinematic equation for linear displacement.
The equation for kinetic energy (K) is given by:
K = (1/2) * m * v^2
where m is the mass of the stone and v is the velocity.
The kinematic equation for linear displacement (d) is given by:
d = v * t
where t is the time.
Given:
Mass of the stone (m) = 1.58 kg
Kinetic energy (K) = 3.11 J
Time (t) = 1.86 s
We can rearrange the equation for kinetic energy to solve for velocity (v):
K = (1/2) * m * v^2
v^2 = (2 * K) / m
v = sqrt((2 * K) / m)
Substituting the given values:
v = sqrt((2 * 3.11 J) / 1.58 kg)
v = sqrt(3.92 m^2/s^2)
v ≈ 1.98 m/s
Now we can use the kinematic equation for linear displacement to find the distance (d) traveled:
d = v * t
d = 1.98 m/s * 1.86 s
d ≈ 3.68 m
Therefore, the stone will go approximately 3.68 meters in 1.86 seconds when moving in a straight line.
To determine how far the stone goes, we can use the equation for kinetic energy and the equation for distance traveled.
The formula for kinetic energy (KE) is given by:
KE = (1/2) * m * v^2
where
m = mass of the object
v = velocity of the object
In this case, the mass of the stone (m) is 1.58 kg, and the kinetic energy (KE) is 3.11 J. We need to determine the velocity (v) of the stone to solve for the distance traveled.
Rearranging the equation for kinetic energy, we can solve for velocity:
v^2 = (2 * KE) / m
v^2 = (2 * 3.11 J) / 1.58 kg
v^2 = 3.94 m^2/s^2
Taking the square root of both sides, we find:
v = √(3.94 m^2/s^2)
v ≈ 1.98 m/s
Now that we have the velocity, we can use the equation for distance traveled:
d = v * t
where
d = distance
v = velocity
t = time
In this case, the time (t) is given as 1.86 s.
So, substituting the given values, we find:
d = 1.98 m/s * 1.86 s
d ≈ 3.68 m
Therefore, the stone goes approximately 3.68 meters in 1.86 seconds.