F = .0566 N still
m = .733
a = F/m = .0772
1.44 = .5 (.0772) t^2
t = 6.11 s
1)What is the net force exerted on the cart-fan combination?
2)Mass is added to the cart until the total mass of the cart-fan combination is 733 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.44 m now? Ignore the effects due to friction.
1)d=a*t^2/2 1.44=(a*4.22^2)/2 a=.1617
F=ma F=.35*.1617 F=.0566N
2) so I got problem 1 right but I'm not sure how to approach/start question 2.
any help is appreciated
m = .733
a = F/m = .0772
1.44 = .5 (.0772) t^2
t = 6.11 s
d = (1/2) a t^2
where d is the distance traveled, a is the acceleration, and t is the time.
Given that the distance (d) is 1.44 m and you want to solve for the time (t), you can rearrange the equation as follows:
t^2 = (2d) / a
Now, you can substitute the values of d and a from the previous problem into the equation:
t^2 = (2 * 1.44) / 0.1617
Simplifying, you get:
t^2 = 17.69
To solve for t, you take the square root of both sides:
t = √17.69
Calculating that, you get:
t ≈ 4.2 s
Therefore, when the mass of the cart-fan combination is increased to 733 g, it takes about 4.2 seconds for the cart to travel 1.44 m starting from rest.
First, we can calculate the acceleration of the cart-fan combination in question 1 using the formula:
a = (2d) / t^2
Using the given values, we can substitute them into the formula to find the acceleration in question 1:
a1 = (2 * 1.44) / (4.22^2)
= 0.1617 m/s^2
Now, let's calculate the force exerted on the cart-fan combination in question 1 using Newton's second law of motion:
F1 = m1 * a1
Substituting the known values, we have:
F1 = (0.35 * 0.1617)
= 0.0566 N
In question 2, the total mass of the cart-fan combination is given as 733 g. Let's convert this mass to kilograms:
m2 = 733 g / 1000
= 0.733 kg
Using Newton's second law, we can find the acceleration of the cart-fan combination in question 2:
a2 = F2 / m2
Since the force remains the same as in question 1, we can use the same force value from question 1:
a2 = F1 / m2
= 0.0566 / 0.733
≈ 0.0772 m/s^2
Now, we can use the formula of motion to find the time it takes for the cart, starting from rest, to travel a distance of 1.44 m in question 2. According to the formula:
d = (1/2) * a2 * t^2
Rearranging the formula, we get:
t^2 = 2d / a2
Substituting the known values:
t^2 = (2 * 1.44) / 0.0772
t^2 ≈ 37.3
t ≈ √(37.3)
≈ 6.1 s
Therefore, it will take approximately 6.1 seconds for the cart, starting from rest, to travel a distance of 1.44 m in question 2 with the increased mass.