50kg iceskater turns a bend at 7m/sec. if the radius of the curve is 5 meters, what is the centripetal force provided by the friction between the blade of the skate and the ice?

Whoever wrote this question does not realize that the force they are asking for is NOT a friction force. It is a lateral force on the EDGE of the blade that keeps the skater on a circular path.

You should mention this to your teacher, if you have one.

If it were a friction force, it would be opposite to the direction of motion and would result in a slowing down of the skater.

The force that are asking for is the centripetal force, M V^2/R

You can do the numbers yourself

To find the centripetal force provided by the friction between the blade of the skate and the ice, we can use the centripetal force equation:

F_c = m * a_c

Where:
F_c is the centripetal force
m is the mass of the ice skater
a_c is the centripetal acceleration

First, we need to calculate the centripetal acceleration. The centripetal acceleration is given by the equation:

a_c = v^2 / r

Where:
v is the velocity of the ice skater
r is the radius of the curve

Plugging in the values, we have:

a_c = (7 m/s)^2 / 5 m
a_c = 49 m^2/s^2 / 5 m
a_c = 9.8 m/s^2

Now that we have the centripetal acceleration, we can calculate the centripetal force:

F_c = 50 kg * 9.8 m/s^2
F_c = 490 N

Therefore, the centripetal force provided by the friction between the blade of the skate and the ice is 490 N.