Question

An 18-kg sled starts up a 36 ∘ incline with a speed of 2.2 m/s . The coefficient of kinetic friction is μk= 0.25.

How far up the incline does the sled travel?

What condition must you put on the coefficient of static friction if the sled is not to get stuck at the point determined in Part A?

If the sled slides back down, what is its speed when it returns to its starting point?

Answer 1

Solution:

Lets use the Energy equations

our equation will write as

where,

• v= initial velocity
• m = mass of sled
• F = friction force
• d = distance friction force acts
• h = the height up the ramp working against the weight or dsin(30)
• F is Fnu or the normal force times the coefficient of friction

Initial kinetic energy = (Potential energy increase) + (work against friction)

Given data

=36⁰

g=9.81m/s²

μ= 0.25

v=2.2m/s

cancel 'm 'on both sides

• coefficient of static friction cannot generate a force greater than mgsin(36)⁰

To make sure it slides back down the incline,the downward weight component must exceed the maximum possible static friction force

coefficient of static friction cannot generate a force greater than mgsin(36)

conservation of energy equations again

find v value

we know that d=0.312m

The sled has kinetic potential while it began the adventure.Of this partly became into misplaced in friction & partly switched over to capacity potential via distinctive function of its place above floor.