The Combination of g on an Incline Lab and the Hooke’s Law Lab

By Felix and Martin

Several weeks ago, we did “g on an Incline” lab and the “Hooke’s Law” lab. From these labs, we are

tempted to use our creativity to combine those experiments into one experiment to find the ΔX of the

spring from two cases of cart on a frictionless slide and slide with friction. From the Hooke’s Law, we

know the formula of F = k x ΔX. K is the spring constant and ΔX is the extension of position. The

extension is calculated from the final position subtracted by the initial position (X2 – X1). From our lab

data, we got the spring constant of spring#1 which is 28.0 N m-1.

From g on an incline, we got a formula of force in the cart, we know W sin Ө = M x a. M is a mass of a

cart and a is the acceleration of the cart. From the data, we know mass of the cart#14 is 503 g. and the

length of the track#1 is 118 cm, and the height of each block is 3.1 cm. We assume that the track and

the cart is made of steel and we are using 4 blocks in this experiment. We also need to convert the mass

into SI units which is kg.

4x 3.1 cm = 12.4 cm

503g x 1 kg

1000𝑔 = 0.503 kg

Sin Ө is calculated by opposite / hypotenuse = 12.4 cm / 118 cm = 0.105. From that, we can find Ө

from arc sin Ө = 6.03.

Cos Ө = Cos 6.03 = 0.994.

First question, if you attach one of the springs from the “Hooke’s Law lab” to the cart on the inclined

track from the “g on an incline” lab, how far down the track will the cart pull the spring.

We calculate ΔX from the data that we have.

Because of the cart has a very small friction, we assume that there is no friction in question number 1.

The acceleration is 0 because the velocity is constant. Thus, we can combine formula I and II together so

Σ Fx = 0. Gravitational force (g) is 9.81 m/s2.

Σ Fx = 0

F – F spring = 0

W sin Ө = 28.0 N m-1 x ΔX

ΔX = 0.503 kg x 9.81 ms−2 𝑥 0.105

28.0 𝑁 𝑚−1 = 0.0185 m = 1.85 cm

Second question, if you turn the cart over so the side without the wheels is against the track, you will

also have to consider friction. If you do this, how much farther down the track can you pull the cart

and still have friction hold it in place?

In the second question we still using the same Hooke’s Law formula which is F = k x ΔX. We must use

the force in the cart = W sin Ө - µK W Cos Ө . Ө is the angle of the cart on an incline.

Because the velocity of the cart is constant the acceleration become 0. There friction is heading down

because when the cart force is going up, so the friction is heading opposite direction. From School for

Champions, the µK of steel on steel is 0.4. Here is the equation between 3 forces.

Σ Fx = 0

F + F friction – F spring = 0

W sin Ө + µK W cos Ө = 28.0 x ΔX

ΔX = (0.503 kg x 9.81 ms−2 𝑥 0.105)+( µK x 0.503 𝑘𝑔 𝑥 9.81 ms−2 𝑥 0.994)

28.0 𝑁 𝑚−1 = 0.0886 m = 8.86 cm

In conclusion, the extension in question one is 1.85cm and the extension in question two which there is

friction is 8.86cm.