a 4 kg block is moving at 12 m/s on a horizontal frictionless surface. a constant force is applied such that the block slows with an acceleration of 3 m/s^2. how much work must this force do to stop the block?
F = ma
work = force * distance
(think of the sliding block as compressing a spring, where work = 1/2 kx^2)
To find the work done by the force to stop the block, we need to calculate the net force acting on the block, and then use the work-energy principle.
First, let's find the net force acting on the block. We can use Newton's second law of motion, which states that the net force (F_net) on an object is equal to its mass (m) multiplied by its acceleration (a):
F_net = m * a
Given that the mass (m) of the block is 4 kg and the acceleration (a) is -3 m/s^2 (negative because the block is slowing down), we can calculate the net force:
F_net = 4 kg * (-3 m/s^2)
= -12 N
Now, we can use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy. Since the block comes to a stop, its final kinetic energy is zero. Therefore, the work done by the force must be equal to the initial kinetic energy of the block:
Work = ΔK
ΔK = K_final - K_initial
Since K_final = 0 (the block comes to a stop), we only need to calculate the initial kinetic energy.
The formula for kinetic energy (K) is given by:
K = (1/2) * m * v^2
Given that the mass (m) of the block is 4 kg and the initial velocity (v) is 12 m/s, we can calculate the initial kinetic energy:
K_initial = (1/2) * 4 kg * (12 m/s)^2
= 288 J
Therefore, the work done by the force to stop the block is equal to the initial kinetic energy:
Work = 288 J
So, the force must do 288 Joules of work to stop the block.