Question

A person on a sled coasts down a hill and then goes over a slight rise with speed 2.7 m/s. The top of this rise can be modeled as a circle of radius 4.1 m. The sled and occupant have a combined mass of 110 kg. If the coefficient of kinetic friction between the snow and the sled is 0.10, what friction force is exerted on the sled by the snow as the sled goes over the top of the rise?

Answer 1

The friction force exerted on the sled by the snow is 88.2 N.

Step-by-step explanation:

Given that,

Speed = 2.7 m/s

Radius = 4.1 m

Combined mass = 110 kg

Coefficient of kinetic friction = 0.10

We need to calculate the normal force

Using newton second law on combine mass

`mg-N=(mv^2)/(r)`

`N=mg-(mv^2)/(r)`

`N=m(g-(v^2)/(r))`

Put the value into the formula

`N=110*(9.8-((2.7)^2)/(4.1))`

`N=882.41\ N`

We need to calculate the friction force is exerted on the sled by the snow

Using formula of frictional force

`f=\mu N`

Put the value into the formula

`f=0.10*882.41`

`f=88.2\ N`

Hence, The friction force exerted on the sled by the snow is 88.2 N.