Question

A 0.62 kg object is at rest. A 2.51 N force to the right acts on the object during a time interval of 1.63 s. a) What is the velocity of the object at the end of this time interval?

Answer 1

Using Newton's second law, the acceleration of the object was found to be 4.05 m/s². The final velocity was then calculated using the kinematic equation, resulting in a velocity of 6.60 m/s to the right at the end of the 1.63 s time interval.

Step-by-step explanation:

To determine the velocity of the object at the end of the time interval, we can use Newton's second law of motion which states that the force acting on an object is equal to the mass of the object times its acceleration (F = ma). From the given problem:

• Mass (m) = 0.62 kg
• Force (F) = 2.51 N
• Time (t) = 1.63 s

We can calculate the acceleration (a) of the object using the formula:

a = F / m

a = 2.51 N / 0.62 kg

a = 4.05 m/s2

Since the object is initially at rest, its initial velocity (u) is 0 m/s. We can use the kinematic equation for velocity:

v = u + at

v = 0 m/s + (4.05 m/s2)(1.63 s)

v = 6.60 m/s

Therefore, the velocity of the 0.62 kg object at the end of the 1.63 s time interval is 6.60 m/s to the right.