Question

On a horizontal, linear track lies a cart that has a fan attached to it. The mass of the cart plus fan is 364 g. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.49 s to travel a distance of 1.43 m. Now, mass is then added to the cart until the total mass of the cart-fan combination is 789 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.43 m? Assume that the cart travels with constant acceleration and ignore the effects due to friction.

Answer 1

Answer:

6.62s

Explanation:

Metric unit conversion:

364 g = 0.364 kg

789 g = 0.789 kg

Starting from rest, the cart takes 4.49 s to travel a distance of 1.43 m. We can use the following equation of motion to calculate the constant acceleration

$s = a_1t_1^2/2$

$a_1 = \frac{2s}{t_1^2} = \frac{2*1.43}{4.49^2} = 0.142 m/s^2$

Using Newton's 2nd law, we can calculate the force generated by the fan to push the 0.364 kg cart forward

$F = a_1m_1 = 0.142*0.364 = 0.052 N$

Now that more mass is added, the new acceleration of the 0.789 kg cart is

$a_2 = F/m_2 = 0.052 / 0.789 = 0.065 m/s^2$

We can reuse the same equation of motion to calculate the time it takes to travel 1.43 m from rest

$s = a_2t_2^2/2$

$t_2^2 = 2s/a_2 = 2*1.43/0.065 = 43.7$

$t_2 = \sqrt{43.7} = 6.62s$

Answer 2

Answer:

6.32 s.

Explanation:

Given:

Mass of cart = 364 g

= 0.364 kg

Mass of cart with added mass = 789 g

= 0.789 kg.

Using equations of motion;

S = ut + 1/2 * at^2

S = a * t^2/2

1.43 = (a * 4.29^2)/2

= 0.155 m/s^2

F = m * a

= 0.364 * 0.155

= 0.0564 N

Acceleration, a2 of the new mass combination = F/m2

= 0.0564/0.789

= 0.0715 m/s^2

Calculating the new time, t2;

S = a * t2^2/2

1.43 = (0.0715 * t2^2)/2

t2^2 = 40

t2 = sqrt(40)

= 6.32 s.

= 0.155 m/s^2