- #1

cutielollipop

- 8

- 0

(mentor note: moved from Classical Physics forum hence no template)

Hello, I am having trouble with this question: Imagine in real life there was a coefficient of kinetic friction of 0.4 between the plastic wheels of the cart and the wooden ramp. If there is only friction on the flat part of the track, how far would the cart go before it stopped if it started from a

height of 4m? The mass of the cart is 10 kg. Here is what I have done and I would appreciate it if someone could verify my work.

Velocity using the simulation at the bottom of ramp: 8.854 m/s

Step 2: Calculating the force of friction on the flat part of the track:

Fk = µkFn

Fn = (5 kg)(9.8 m/s^2)

Fn = 49 N

Fk = (0.4)(49 N)

Fk = 19.6 N

Step 3: Calculating distance before the cart stops.

The work done by friction:

Wfriction = Ffriction x distance

The work done by friction is also equal to the loss in kinetic energy:

Wfriction = Ek_initial – Ek_final

Wfriction = ½ mv^2 – 0 (because Ek_final = 0)

Equating expressions:

Ffriction x d = ½ mv^2

Rearranging for “d”

d = 10.0 m

Therefore, the car would go 10 m before it stops on the flat part of the track.

Hello, I am having trouble with this question: Imagine in real life there was a coefficient of kinetic friction of 0.4 between the plastic wheels of the cart and the wooden ramp. If there is only friction on the flat part of the track, how far would the cart go before it stopped if it started from a

height of 4m? The mass of the cart is 10 kg. Here is what I have done and I would appreciate it if someone could verify my work.

Velocity using the simulation at the bottom of ramp: 8.854 m/s

Step 2: Calculating the force of friction on the flat part of the track:

Fk = µkFn

Fn = (5 kg)(9.8 m/s^2)

Fn = 49 N

Fk = (0.4)(49 N)

Fk = 19.6 N

Step 3: Calculating distance before the cart stops.

The work done by friction:

Wfriction = Ffriction x distance

The work done by friction is also equal to the loss in kinetic energy:

Wfriction = Ek_initial – Ek_final

Wfriction = ½ mv^2 – 0 (because Ek_final = 0)

Equating expressions:

Ffriction x d = ½ mv^2

Rearranging for “d”

d = 10.0 m

Therefore, the car would go 10 m before it stops on the flat part of the track.

Last edited by a moderator: