A 0.50-KG BLOCK INITIANALLY ON A REST ON A FRICTIONLESS HORIZONTAL SURFACE IS ACTED UPON BY A FORCE OF 8.0N FOR A DISTANCE OF 4.0M. HOW MUCH KINETIC ENERGY DOES THE BLOCK GAIN?
work done is forceapplied*distance, since no work is absorbed by friction, the final KE must be....
32 J
To calculate the kinetic energy gained by the block, we need to use the formula for kinetic energy.
The formula for kinetic energy is given by:
Kinetic Energy = 0.5 * m * v^2
where:
m is the mass of the object (in kilograms)
v is the velocity of the object (in meters per second)
In this case, the block has a mass of 0.50 kg, and we need to find the velocity to calculate the kinetic energy gained.
To find the velocity, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration:
Net Force = m * a
In this case, the net force is 8.0 N (given in the problem), and the mass of the block is 0.50 kg. We need to find the acceleration to calculate the velocity.
We can use the following equation to find the acceleration:
Net Force = m * a
8.0 N = 0.50 kg * a
Solving for a, we get:
a = 8.0 N / 0.50 kg
a = 16 m/s^2
Now that we have the acceleration, we can find the final velocity using the following equation:
v^2 = u^2 + 2 * a * d
where:
u is the initial velocity (which is zero in this case)
a is the acceleration (which we just found to be 16 m/s^2)
d is the distance (which is given as 4.0 m)
Plugging in the values, we get:
v^2 = 0^2 + 2 * 16 m/s^2 * 4.0 m
v^2 = 128 m^2/s^2
To find the velocity, we take the square root of both sides:
v = sqrt(128 m^2/s^2)
v = 11.31 m/s (approximately)
Now that we have the velocity, we can calculate the kinetic energy using the formula mentioned earlier:
Kinetic Energy = 0.5 * 0.50 kg * (11.31 m/s)^2
Calculating the kinetic energy:
Kinetic Energy = 0.5 * 0.50 kg * (127.6 m^2/s^2)
Kinetic Energy = 31.9 Joules
Therefore, the block gains 31.9 Joules of kinetic energy.