A 0.5 kg block is sliding along a table top with an initial velocity of 0.2 m/s. It slides to rest in a distance of 70 cm. Find the frictional force that is slowing its motion.
KE=W(fr)
KE = m•v²/2
W(fr) =F(fr) •s
m•v²/2 =F(fr) •s
F(fr)= m•v²/2•s
142.85
To find the frictional force that is slowing down the block's motion, we need to use the equation of motion for accelerated motion.
First, let's determine the acceleration of the block. We can use the equation:
(vf^2) = (vi^2) + 2ad
Where:
- vf is the final velocity (which is 0 m/s since the block comes to rest)
- vi is the initial velocity (0.2 m/s)
- a is the acceleration
- d is the distance covered (70 cm or 0.7 m)
Rearranging the equation, we have:
a = (vf^2 - vi^2) / (2d)
Substituting the given values, we get:
a = (0 - (0.2^2)) / (2 * 0.7)
a = -0.04 / 1.4
a = -0.0286 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the initial velocity (i.e., it is deceleration).
Now, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration:
F = m * a
Substituting the given mass of the block (0.5 kg) and the calculated acceleration (-0.0286 m/s^2), we have:
F = 0.5 kg * (-0.0286 m/s^2)
F = -0.0143 N
The negative sign indicates that the frictional force is in the opposite direction of the initial motion of the block. Therefore, the frictional force that is slowing down the block's motion is approximately 0.0143 Newtons.