Question

A 3.0 kg rock is dropped from rest. What will its velocity be in 6.0 s if a 7.0 N force of air resistance acts on it?

Answer 1

Answer:

14 m/s^2

Explanation:

$v_{f}=V_{i} +at$

$V_{f}=$final velocity

$V_{i}=$initial velocity

a= acceleration

t=time

Initial velocity is zero, since the rock was originally at rest and time is 6 seconds.

You are looking for the final velocity, but you need acceleration to find it. Remember that:

$F=m*a$

Where F is force, m is mass, and a is acceleration. You know force is 7.0 N and mass is 3.0 kg. Use these to find a.

$7.0 N= 3*a$

$\frac{7}{3} = a$

So now...

$v_{f}=V_{i} +at$

$v_{f}=0 +(\frac{7}{3}) (6)$

$v_{f}= 14 m/s^{2}$

Answer 2

Explanation:

F=ma

m=3kg

F=7N

7=3a

a=7/3

acceleration=velocity/time taken

7/3=velocity/6

velocity=7/3×6/1

velocity=14ms-¹