Question

question 6.
a student is pulling a 100 - kilogram cart, accelerating at 1.5 meters per second squared. a second student joins the first student pulling in pulling the cart in the same direction, the carts acceleration increases to 2.5 m/s².
with how much force is the second student pulling?
a. 50 n
b. 100 n
c. 150 n
d. 250 n

Explanation:

Step1: Calculate the initial force

Using Newton's second law \(F = ma\), where \(m = 100\space kg\) and \(a_1=1.5\space m/s^{2}\).
\(F_1=100\times1.5 = 150\space N\)

Step2: Calculate the final force

Using \(F = ma\) again, with \(a_2 = 2.5\space m/s^{2}\)
\(F_2=100\times2.5=250\space N\)

Step3: Find the force of the second student

The force of the second student \(F_{second}=F_2 - F_1\)
\(F_{second}=250 - 150=100\space N\)

Answer:

B. \(100\space N\)