Question

.125 kg piece of rubber slows down from 20 m/s to 10m/s as it slides along ice for 102 meters calculate the force of friction

Answer 1

Answer:

-0.18 N

Explanation:

The motion of the piece of rubber is a uniformly accelerated motion, so we can find its acceleration using the suvat equation:

$v^2-u^2=2as$

where:

v = 10 m/s is the final speed

u = 20 m/s is the initial speed

a is the acceleration

s = 102 m is the distance  covered

Solving for a,

$a=\frac{v^2-u^2}{2s}=\frac{10^2-20^2}{2(102)}=-1.47 m/s^2$

The net force acting on the piece of rubber is the force of friction; according to Newton's second law of motion, the force is equal to the product of mass (m) and acceleration (a):

$F=ma$

Here we have

m = 0.125 kg is the mass

Therefore, the force of friction is:

$F=ma=(0.125)(-1.47)=-0.18 N$

And the negative sign means the direction of the force is opposite to the direction of motion.