A .5 kg block is sliding along a tabletop with an initial velocity of .20 m/s. It slides to rest in a distance of .7m. Find the frictional force regarding it motion.
The answer given is .0143 N but I do not understand how to get that.
Frictional Force=1/2mv^2/d
=1/2(0.5)(0.2)^2/0.7
=1/2(0.5)(0.04)/0.7
=1/2(0.02)/0.7
=0.01/0.7
=0.0142857143
=0.0143 N
To find the frictional force acting on the block, you can use the following equation:
frictional force = mass × deceleration
First, calculate the deceleration of the block using the given information. The block starts with an initial velocity and comes to rest, so the final velocity is zero. You can use the following equation of motion to find the deceleration:
final velocity^2 = initial velocity^2 + 2 × acceleration × distance
Plugging in the given values:
- Final velocity (vf) = 0 m/s
- Initial velocity (vi) = 0.20 m/s
- Distance (d) = 0.7 m
0 = (0.20 m/s)^2 + 2 × acceleration × 0.7 m
0 = 0.04 m^2/s^2 + 1.4 m/s^2 × acceleration
Simplifying the equation, we get:
1.4 m/s^2 × acceleration = -0.04 m^2/s^2
Acceleration = -0.04 m^2/s^2 ÷ 1.4 m/s^2
Acceleration = -0.0286 m/s^2
Now that we have the acceleration, we can determine the frictional force by multiplying the mass of the block by the deceleration:
frictional force = mass × deceleration
frictional force = 0.5 kg × -0.0286 m/s^2
frictional force ≈ -0.0143 N
Note that the negative sign indicates that the frictional force is acting in the opposite direction of the motion. In this case, it opposes the motion of the block. The magnitude of the force is 0.0143 N, as given in the answer.