Question

In Physics lab, a lab team places a cart on one of the horizontal, linear tracks with a fan attached to it. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.34 s to travel a distance of 1.62 m. The mass of the cart plus fan is 354 g. Assume that the cart travels with constant acceleration.

A) What is the net force exerted on the cart-fan combination?B) Mass is added to the cart until the total mass of the cart-fan combination is 762 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.62 m now?

Answer 1

A. F = 0.06 N

B. t = 6.37 s

Step-by-step explanation:

A)

First we need to find the constant acceleration of the cart. For this purpose, we use 2nd equation of motion:

s = (Vi)(t) + (0.5)at²

where,

s = distance traveled = 1.62 m

Vi = 0 m/s (Since, it starts from rest)

t = Time Taken = 4.34 s

a = acceleration = ?

Therefore,

1.62 m = (0 m/s)(4.34 s) + (0.5)(a)(4.34 s)²

1.62 m/9.4178 s² = a

a = 0.172 m/s²

Now, from Newton's Second law, we know that:

F = ma

where,

F = Net Force of the combination = ?

m = Mass pf combination = 354 g = 0.354 kg

Therefore,

F = (0.354 kg)(0.172 m/s²)

F = 0.06 N

B)

Now, for the same force, but changed mass = 762 g = 0.762 kg, we have the acceleration to be:

F = ma

a = F/m

a = 0.06 N/0.762 kg

a = 0.08 m/s²

Now, using 2nd equation of motion:

s = (Vi)(t) + (0.5)at²

1.62 m = (0 m/s)(t) + (0.5)(0.08 m/s²)t²

t² = 1.62 m/(0.04 m/s²)

t = √40.54 s²

t = 6.37 s