Question

A 0.50 kg object is at rest. A 2.88 N force to the right acts on the object during a time interval of 1.48 s. a) What is the velocity of the object at the end of this time interval? Answer in units of m/s.

Answer 1

Answer:

8.5m/s

Explanation:

We are given that

Mass of object=m=0.50 kg

Initial velocity, u=0

Force=F=2.88 N

Time=1.48 s

a.We know that

$F=ma$

Using the formula

$2.88=0.50a$

$a=\frac{2.88}{0.50}=5.76m/s^2$

$a=\frac{v-u}{t}$

Using the formula

$5.76=\frac{v-0}{1.48}$

$v=5.76\times 1.48=8.5m/s$

Hence, the velocity of the object at the end of this time interval=8.5m/s